NCERT Solutions Class 12 Physics

NCERT Solutions for class 12 Physics 

The National Council of Educational Research and Training (NCERT) publishes Physics textbooks for Class 12. The NCERT Class 12th Physics textbooks are well known for their up-to-date and thoroughly revised syllabus. The NCERT Physics Books are based on the most recent exam pattern and CBSE syllabus.

NCERT keeps the Physics books up-to-date by using the most recent year’s question papers. The NCERT Class 12 Physics books are well-known for their presentation. The use of NCERT Books Class 12 Physics is not only appropriate for studying the regular syllabus of various boards, but it can also be beneficial for candidates taking various competitive exams, such as Engineering Entrance Exams, and Olympiads.

NCERT Solutions for class 12 Physics – CBSE 2022-23 Download for Free

Students will benefit greatly from NCERT Solutions for Class 12 Physics as they prepare for their board exams. The solution sets are created by an Extramarks team of experts who are familiar with the NCERT marking scheme and exam format. This solution set will assist students in understanding the concepts covered in this chapter and in preparing for their boards and entrance exams. Scroll down to find the link to the solution set for free download.

NCERT Solutions for class 12 physics – Download 

The NCERT Solutions for 12 Physics subjects are available here. Students can access the solutions by clicking on the links for the chapters in which they are interested. All of the questions are answered here in relation to the methods and procedures described in the textbook. All of the chapters’ are available here for students to download and study offline.

NCERT Solutions Exercise for class 12 physics

There are 15 chapters in total for the Class 12 Physics syllabus. It is further divided into 2 terms. The syllabus for Term 1 includes chapters 1, 2, 3, 4, 5, 6, and 7, while Term 2 has chapters 8, 9, 10, 11, 12, 13, 14, and 15. Students can check the chapter and download the solution for that topic.

Physics Class 12: Chapter 1 Electric Charges and Fields

This is the first chapter of the physics textbook for Class 12. Students will learn about electric fields, charges, and their functional areas in this chapter. You will also learn about related laws and theorems, such as Gauss’ Law. Some may find this chapter difficult, but with Class 12th Physics NCERT solutions, understanding and answering the questions will be a breeze.

This chapter will assist you in answering questions based on the calculation of forces between two charged particles placed at a distance from one another. There are also questions about calculating the total charge and dipole moment of a system with torque and an electric field.

Topics covered in Term 1 from Chapter 1

Conservation of Charge, Coulomb’s law-force between two-point charges, Electric Charges, forces between multiple charges; superposition principle and continuous charge distribution. Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field due to a dipole, torque on a dipole in uniform electric field. Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet.

Physics Class 12: Chapter 2 Electrostatic Potential and Capacitance

This is the second chapter of the physics textbook for students in Grade 12. The potential difference and the development of electrostatic potential will be covered in the second chapter. This chapter also contains important formulas and derivations that you must use to answer the numerical questions. The electrostatic potential will be introduced to students in a nutshell.

This chapter includes derivations for topics such as the potential energy of a charge system, energy stored in a capacitor, and potential due to an electric dipole. According to the CBSE syllabus, problems based on the effective capacitance of capacitors are also solved step-by-step in the NCERT Solutions for Class 12 Physics. These concepts and numericals are answered in the simplest way possible to assist students in performing well on the board exams. 

Topics covered in Term 1 from Chapter 2

Electric potential, potential difference, electric potential due to a point charge, a dipole and system of charges; equipotential surfaces, electrical potential energy of a system of two-point charges and of electric dipole in an electrostatic field. Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics and electric polarisation, capacitors and capacitance, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor.

Physics Class 12: Chapter 3 Current Electricity

Students will learn the definitions of key terms such as conductance, resistance, cell EMF, drift velocity, conductance, resistivity, and internal resistance in this chapter. The graphing of resistivity for semiconductors, metals, and alloys will also be taught to students. It will give you a firm understanding of converting complex circuits into simple series and parallel circuits. This chapter provides a brief explanation of the concepts of metre bridges and potentiometers. At the end of the chapter, you can practise a variety of problems based on these concepts.

To solve the problems in this chapter, students must pay attention while learning the derivations of each law. If students have any questions while solving, they can refer to the NCERT Solutions for Class 12th Physics solved by Extramarks experts, which can be found at the link provided below.

Topics covered in Term 1 from Chapter 3

Electric current, flow of electric charges in a metallic conductor, drift velocity, mobility and their relation with electric current; Ohm’s law, electrical resistance, V-I characteristics (linear and nonlinear), electrical energy and power, electrical resistivity and conductivity; temperature dependence of resistance. Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series and in parallel, Kirchhoff’s laws and simple applications, Wheatstone bridge, metre bridge(qualitative ideas only). Potentiometer – principle and its applications to measure potential difference and for comparing EMF of two cells; measurement of internal resistance of a cell (qualitative ideas only)

Physics Class 12: Chapter 4 Moving Charges and Magnetism 

This is the fourth chapter of the Class 12 Physics textbook. This chapter will help you understand the laws of electricity and magnetism. You’ll also see how the magnetic field exerts forces on moving charged particles like electrons, protons, and current-carrying wires. This is an important chapter not only for the Class 12 exams, but also for further studies. This chapter will lay a solid foundation for students who want to ace competitive examinations. Students will look at a variety of diagrams, examples, graphs, and illustrations from everyday life. By relating it this way, you can easily remember concepts. This chapter also introduces concepts such as forces between two current-carrying wires and Ampere’s Circuital Law. Numerous problems and questions based on these concepts are provided at the end of the chapter. Students should practice these questions on a regular basis in order to gain a thorough understanding of the chapter.

Topics covered in Term 1 from Chapter 4

Concept of magnetic field, Oersted’s experiment. Biot – Savart law and its application to the current carrying circular loop. Ampere’s law and its applications to infinitely long straight wire. Straight and toroidal solenoids (only qualitative treatment), force on a moving charge in uniform magnetic and electric fields. Force on a current-carrying conductor in a uniform magnetic field, force between two parallel current-carrying conductors-definition of ampere, torque experienced by a current loop in uniform magnetic field; moving coil galvanometer-its current sensitivity and conversion to ammeter and voltmeter.

Physics Class 12: Chapter 5 Magnetism and Matter

This is the fifth chapter of the physics textbook for Class 12. Starting with the bar magnet, you will encounter magnetic field lines, magnetic moment, magnetic field strength, susceptibility, and other concepts. Students will find questions on magnetic field lines, the earth’s magnetism, and compass direction in Chapter 5. In addition, students can learn about interstellar space, its weak magnetic field, and a solenoid’s magnetic moment. 

However, in order to understand these concepts, you must first understand the previous chapters. This chapter requires a thorough understanding of magnetism, with its laws and their mathematical derivations. Candidates studying this chapter must take notes in each section. This, in turn, will aid in answering the chapter’s questions. In case of any doubts, students can consult the Class 12th Physics NCERT solutions.

Topics covered in Term 1 from Chapter 5

Current loop as a magnetic dipole and its magnetic dipole moment, magnetic dipole moment of a revolving electron, bar magnet as an equivalent solenoid, magnetic field lines; earth’s magnetic field and magnetic elements.

Physics Class 12: Chapter 6 Electromagnetic Induction

This is the sixth chapter of the physics textbook for students in Grade 12. Magnetism and electricity are explained in relation to one another in chapter 6. Lenz’s Law and Faraday’s Law are two of the most important ideas in this chapter. This chapter also contains questions and numerical problems based on this concept for students’ use. They must answer these questions in order to gain a thorough understanding of the subject matter. While answering these questions, they should keep in mind the magnetic field and the direction of induced EMF.

Topics covered in Term 1 from Chapter 6

Electromagnetic induction; Faraday’s laws, induced EMF and current; Lenz’s Law, Eddy currents. Self and mutual induction.

Physics Class 12: Chapter 7 Alternating Current

Students will learn about applying an alternating current voltage to an inductor, a series LCR circuit, a resistor and capacitor, and a transformer in this chapter. The chapter includes a number of graphs and diagrams to help students fully grasp the concepts. Students will undoubtedly be prepared to solve the difficult numerical problems as well as questions from this chapter on the exam if they practise regularly. This chapter is essential for both board and competitive exams. As a result, if students pay close attention to this chapter, they should do well on both exams.

Topics covered in Term 1 from Chapter 7 

Alternating currents, peak and RMS value of alternating current/voltage; reactance and impedance; LC oscillations (qualitative treatment only), LCR series circuit, resonance; power in AC circuits. AC generator and transformer.

Physics Class 12: Chapter 8 Electromagnetic Waves

This is the eighth chapter of the physics textbook for students in Grade 12. Students will learn about electromagnetism, capacitors, potential differences, and displacement in detail in this chapter. Kirchhoff’s rules, the RMS value of conduction current, and the similarities between conduction current and displacement current are also discussed in this chapter.

The topic of magnetism also applies to electromagnetic waves, and your previous knowledge will help you master this chapter. However, if you are having trouble understanding the material, the NCERT Solutions for 12 Physics are available for your reference

Topics covered in Term 2 from Chapter 8

Electromagnetic waves, their characteristics, their Transverse nature (qualitative ideas only). Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays) including elementary facts about their uses.

Physics Class 12: Chapter 9 Ray Optics and Optical Instruments

Students will learn about light refraction, reflection, and dispersion using the ray diagram in this chapter. The formation of images is clearly explained in this chapter using the laws of reflection and refraction. This chapter discusses the design and operation of optical instruments, including the human eye. Students will also learn about natural phenomena caused by sunlight, which will be useful in exams.

Topics covered in Term 2 from Chapter 9

Ray Optics: Refraction of light, total internal reflection and its applications, optical fibres, refraction at spherical surfaces, lenses, thin lens formula, lensmaker’s formula, magnification, power of a lens, combination of thin lenses in contact, refraction of light through a prism.

Optical instruments: Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers

Physics Class 12: Chapter 10 Wave Optics

Huygen’s laws of reflection and refraction are the most important concept in Chapter 12 and has the highest weightage of marks. Brewster’s Law, the Doppler Effect, and angular width expression are among the other concepts covered in this chapter. For students’ understanding, these are discussed with appropriate examples. Students should go over all of the derivations and theories in this chapter, as well as solve all of the numerical problems and questions.

Topics covered in Term 2 from Chapter 10

Wave optics: Wavefront and Huygens principle, reflection and refraction of plane waves at a plane surface using wave fronts. Proof of laws of reflection and refraction using Huygens principle. Interference, Young’s double slit experiment and expression for fringe width, coherent sources and sustained interference of light, diffraction due to a single slit, width of central maximum.

Physics Class 12: Chapter 11 Dual Nature of Radiation and Matter

This is the eleventh chapter of the physics textbook for Class 12. In this section, you will learn whether radiation is similar to a particle, a wave, or both in properties and appearance. It also includes concepts such as photoelectric effects, wave theory, wave nature, and so on.

This chapter discusses important topics like the photoelectric effect, Einstein’s Photoelectric equation, electron emission, the wave aspect of matter, and experimental effects. This chapter’s numerical problems should be studied more carefully by students. They must read all of the chapters and answer the textbook’s questions to secure a higher exam score.

Topics covered in Term 2 from Chapter 11

Dual nature of radiation, Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation-particle nature of light. Experimental study of photoelectric effect. Matter waves-wave nature of particles, de-Broglie relation.

Physics Class 12: Chapter 12 Atoms

The electron in the classical model of an atom revolves around the nucleus like a planet around the sun. The key concept in this chapter is the Bohr Model of the Hydrogen Atom. Because it contains many problems, only by learning and revising concepts on a regular basis will students be able to score well.To achieve good grades, it is necessary to memorise the important formulas and steps to be taken when solving problems. Students can use the NCERT Solutions for Physics Class 12 to see how these formulas are used and try them out for themselves..

Topics covered in Term 2 from Chapter 12

Alpha-particle scattering experiment; Rutherford’s model of atom; Bohr model, energy levels, hydrogen spectrum.

Physics Class 12: Chapter 13 Nuclei

Students will learn everything there is to know about the Nucleus in this chapter. You will discover that the density of nuclear matter is independent of nucleus size, whereas the mass density of an atom does not follow this rule. Nucleus composition, mass-energy, nuclear force, nuclear energy, atomic masses, nucleus size, nuclear binding energy, and radioactivity are all topics covered in Chapter 13.

Topics covered in Term 2 from Chapter 13

Nuclei Composition and size of nucleus, Nuclear force, Mass-energy relation, mass defect, nuclear fission, nuclear fusion.

Physics Class 12: Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits

This is the fourteenth chapter of the physics textbook for Class 12. This unit discusses semiconductors, their various types, insulators, and applications. It is one of the most highly rated chapters in the Class 12 exams. The majority of the questions in this chapter will revolve around n-type semiconductors, dopants, and carries. This chapter’s exam may also include true or false questions about germanium, carbon, and silicon semiconductors. Another important concept in this chapter is how to calculate the energy gap. Students should thoroughly learn how to calculate the output frequency of a half-wave and full-wave rectifier, as questions on this topic will be asked. This chapter includes several additional exercises in which you must determine the number of electrons and holes in arsenic, indium, and silicon atoms.

Topics covered in Term 2 from Chapter 14

Energy bands in conductors, semiconductors and insulators (qualitative ideas only) Semiconductor diode – I-V characteristics in forward and reverse bias, diode as a rectifier; Special purpose p-n junction diodes: LED, photodiode, solar cell.

Physics Class 12: Chapter 15 Communication Systems

This is the fifteenth chapter of the Class 12 Physics textbook. The final chapter will go over communication systems and their components in detail. Communication systems, classification of communication systems into Technology, Media, and Application Area, and modern electronics components such as sensors, emitters, transducers, and amplifiers are discussed in this chapter.

Marking Scheme for CBSE 2022-23

Term I

Unit No: Name of Unit Marks
Unit – I Electrostatics 17
Chapter-1: Electric Charges and Fields
Chapter-2: Electrostatic Potential and Capacitance
Unit – II Current Electricity
Chapter-3: Current Electricity
Unit – III Magnetic Effects of Current and Magnetism 18
Chapter-4: Moving Charges and Magnetism
Chapter-5: Magnetism and Matter
Unit – IV Electromagnetic Induction and Alternating Currents
Chapter-6: Electromagnetic Induction
Chapter-7: Alternating Currents
Total 35

Term II

Unit No: Name of Unit Marks
Unit – V Electromagnetic Waves 17
Chapter-8: Electromagnetic Waves
Unit – VI Optics
Chapter-9: Ray Optics and Optical Instruments
Chapter-10: Wave Optics
Unit – VII Dual Nature of Radiation and Matter 11
Chapter-11: Dual Nature of Radiation and Matter
Unit – VIII Atoms and Nuclei
Chapter-12: Atoms
Chapter-13: Nuclei
Unit – IX  Electronic Devices 7
Chapter-14: Semiconductor – Electronics: Materials, Devices and Simple Circuits
Total 35

Key specifications of Extramarks’s NCERT Solutions for class 12 physics

  • All numerical problems mentioned in the textbooks are explained step-by-step in the solutions book. Students can practise these sums by seeing how the formula work in that problem
  • All logical and theoretical problems are explained in detail. Students can read through the solutions and create their own short notes for revision
  • The solutions contain answers that are curated and verified by subject matter experts. These questions are therefore, focused on important topics and have detailed explanations
  • A bonus is that there are free, unlimited downloads for students to access all relevant study materials

Why study NCERT solutions for class 12 physics?

If you want to pursue a career in engineering or a related field, you must excel in physics. This is where NCERT Solutions Class 12 Physics can help. These NCERT Solutions for Class 12 Physics help students prepare for board exams as well as engineering/medical entrance exams. It provides concrete knowledge and assists students in comprehending all of the important concepts covered in each chapter of the NCERT textbooks.

  • Answers that are simple and precise
  • Curated by well-known academics
  • Covers both complex and simple topics

Why do toppers give priority to Extramarks NCERT Solutions?

Extramarks offers one of the most student-friendly solutions available on the internet. Students can now easily complete the NCERT Class 12 Physics Syllabus by using these solutions. The following are some of the key features of the NCERT Solutions.

  • The solutions are created by highly knowledgeable teachers

Physics is a subject with many concepts that have applications in our daily lives. Students are strongly advised to consult the best study material available online in order to comprehend these concepts. The Class 12th Physics NCERT solutions are designed with the understanding ability of Class 12 students in mind. As a result, the solutions are dependable and precise, as they are entirely based on the CBSE board’s exam pattern.

  • Provides a variety of practice problems

When it comes to final exam preparation, the only way to achieve a high score is to practice. The chapter-by-chapter solutions can be accessed online at the convenience of the student. The solutions are highly authentic, and students can rely on them to help them prepare for their board exams.

  • It is available at all times and from any location

A student’s life is marked by the completion of the Class 12 exam. So, it is critical to select the best study material to meet all of their requirements. When it comes to board exam preparation, Extramarks has developed NCERT Solutions for Physics Class 12 to help students gain confidence in answering complex questions. They offer free solutions that can be used anywhere and at any time while answering the textbook’s exercise questions.

Q.1 When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Ans.

When a glass rod is rubbed with a silk cloth, charges of equal magnitude but of opposite nature appear on the two bodiesas the charges are created in pairs. The algebraic sum of charges produced on the two rubbed bodies is zero. The net charge on the two bodies before rubbing was also zero. Therefore, this phenomenon is consistent with the law of conservation of energy. The similar phenomenon is observed with many other pairs of bodies.

Q.2 A parallel plate capacitor (shown in figure) made of circular plates each of radius R = 6 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V a.c supply with an angular frequency of 300 rads-1.
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of B at a point 3 cm from the axis between the plates.


Ans.

aHere, V rms = 230 V , C =100 pF = 10-10 F , ω = 300 rad s -1As, I rms = V rmsX c = V rmsC ωThus, I rms = 230 V × 10-10 F × 300 rad s -1 = 6.9 µA bYes, the conduction current is equal to the displacement current becauseId = ε0dt= ε0dEAdt Thus, Id = ε0A dVd dt = C dVdtAs,dVdt = ICCThus,Id = CICC= ICconduction current

cAs per the figure given below, using Ampere’s circuital law we get,

B = ε0 µ0 rdE2 dt As, E = qA ε0Thus, dEdt= dqdtA ε0 dEdt= I rmsπ R2 ε0 Hence, B becomes; B = ε0 µ 0r2 x I rmsπ R 2 ε 0 = µ0 I rms r2 π R2Therefore, amplitude of magnetic field will be; B0 = µ0 I0 r2 π R2or, B0 = µ02 I rms2 π R2 B0 = 4 π ×10-7 NA-2 3×10-2 m 1.414 6.9 ×10-6 A2 π ×6 ×10-2 m2 B0 = 1.63 ×10-11 Tesla

Q.3 What physical quantity is the same for X-rays of wavelength 10-10 m, red light of wavelength 6800 Angstrom and radio waves of wavelength 500 m?

Ans.

The speed of light (i.e.) c = 3 × 10 8 ms -1 remains the same in a vacuum for given X rays, red light and radio waves.

Q.4 Four point charges qA = 2 μC, qB = −5 μC, qC = 2 μC, and qD = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

Ans.

Inthe given figure,thereisasquare of side 10 cm with four charges placed at its corners. S is the centre of the square. where, Lengthofsideofsquare=AB=BC=CD=AD= 10 cm Lengthofdiagonalofsquare=AC=BD=10 2 cm AS=SC=DS=SB=5 2 cm A charge of amount 1μC is placed at thecentreSof thesquare. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@425C@

Force of repulsion on 1 μC charge because of 2 μC charges at A and C are equal in magnitude but opposite in direction. Therefore, they cancel each other.
Similarly, force of attraction on 1 μC charge because of -5 μC charges at A and C are equal in magnitude but opposite in direction. Therefore, they also cancel each other.
Therefore, net force on 1 μC charge at centre S is zero.

Q.5 (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
(b) Explain why two field lines never cross each other at any point?

Ans.

(a) An electrostatic field line is a continuous curve because it represents the actual path of the unit positive charge which experiences a continuous force in an electrostatic field. It cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.
(b) If two field lines cross each other at a point, then electric field will show two directions at that point. This is impossible. Therefore, two field lines never cross each other at any point.

Q.6 A plane electromagnetic wave travels in vacuum along z direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?

Ans.

The electric field vector E and magnetic field vector B are perpendicular to each other and they lie in x-y plane.

Ifthefrequencyofwave,ν=30MHzλ=cν=3×108 ms1 30×106 Hz=10 m

Q.7 Two point charges qA = 3 μC and qB = −3 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 × 10−9 C is placed at this point, what is the force experienced by the test charge?

Ans.

( a ) The situationoftheproblem is shown in the given figure. Here,O is the midpoint of line AB. q A =3μC =3×10 -6 C q B =-3μC =-3×10 -6 C Distance between two charges, AB=20 cm AO=OB = 10 cm Net electric field at O=E Electric field at point O caused by +3μC chargeis givenas: E 1 = q A 4π ε 0 ( AO ) 2 = 3× 10 6 4π ε 0 ( 10× 10 2 ) 2 N C 1 alongOB Where, ε 0 = Permittivity of free space 1 4π ε 0 =9× 10 9 N m 2 C 2 Magnitude of electric field at point O caused by 3μC chargeisgivenas: E 2 =| q B 4π ε 0 ( OB ) 2 | =| 3× 10 6 4π ε 0 ( 10× 10 2 ) 2 | = 3× 10 6 4π ε 0 ( 10× 10 2 ) 2 N C 1 alongOB E= E 1 + E 2 = 3× 10 6 4π ε 0 ( 10× 10 2 ) 2 + 3× 10 6 4π ε 0 ( 10× 10 2 ) 2 =2×[ ( 9× 10 9 )× 3× 10 6 ( 10× 10 2 ) 2 ] =5.4× 10 6 N C 1 along OB E 2 = 5.4 × 1 0 6 NC -1 along OB Theelectric field at midpoint O is 5.4×1 0 6 NC 1 along OB. ( b ) A test charge of magnitude 1.5×1 0 9 C is placed atmidpoint O. Here,q=1.5 × 1 0 9 C Letforce experienced by the test charge = F F=qE = 1 .5×10 -9 ×5 .4×10 6 = 8 .1×10 -3 N;along BA. 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Q.8 A radio can tune into any station in the 7.5 MHz to 12 MHz. What is the corresponding wavelength band?

Ans.

As,λ = cν = 3 ×108 ms17.5 × 106 Hz= 40 mSimilarly λ= cν = 3 ×108 ms1 12 ×106 Hz = 25 mThus, the corresponding wavelength band is 40 m to 25 m.

Q.9 A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

Ans.

Frequency of the electromagnetic waves produced by the oscillator will be same as the frequency of oscillating charged particle (i.e.) 109 Hz.

Q.10 A system has two charges qA = 2.5 × 10−7 C and qB = −2.5 × 10−7 C located at points A: (0, 0, − 15 cm) and B: (0, 0, + 15 cm), respectively. What are the total charge and electric dipole moment of the system?

Ans.

The charges q A and q B can be located in a coordinate frame of reference as shown in the given figure. Magnitude of chargeat A, q A = 2 .5 × 10 -7 C Magnitude of chargeat B, q B = -2 .5 × 10 -7 C Total charge of the system, q = q A + q B = 2 .5 × 10 7 C – 2 .5 × 10 -7 C = 0C Distance between thetwo charges,d=15+15 =30 cm =0.3 m Electric dipole moment of the system is given as: p= q A ×d = q B ×d = 2.5×1 0 7 ×0.3 = 7.5×1 0 8 C m along positive zaxis Electric dipole moment of the system is7.5×1 0 8 Cm along positivezaxis. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@EB26@

Q.11 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of electric field part of the wave?

Ans.

As,c=E0B0 thus E0=cB0 =3×108 ms-1×510×109 T =153 NC-1

Q.12 An electric dipole with dipole moment 4 × 10−9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 104 N C−1. Calculate the magnitude of the torque acting on the dipole.

Ans.

Here, electric dipole moment, p = 4×10 -9 Cm Angle made by p with theuniform electric field, θ=30° Magnitudeof electric field, E = 5×10 4 NC -1 Torque acting on the dipole is given by, τ = pE sinθ =4×10 -9 × 5×10 4 sin30° =20×10 -5 × 1 2 =10 -4 Nm Themagnitude of the torque acting on the dipole is 10 -4 Nm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@34FA@

Q.13 A polythene piece rubbed with wool is found to have a negative charge of 3 × 10−7 C.
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene?

Ans.

( a )Here, charge on the polythene piece, q=-3×10 -7 C Charge on an electron, e = -1 .6×10 -19 C Letnumber of electrons transferred to polythene piecefromwool= n Wecancalculatenby using the relation, q = ne n= q e = -3×10 -7 -1 .6×10 -19 = 1 .87×10 12 Number of electrons transferred from wool to polythene is 1 .87 ×10 12 . ( b ) Yes,transfer of mass takes placefromwoolto thepolythenepiece. Asmassofelectron, m e = 9 .1×10 -31 kg Total mass transferred fromwooltothepolythene,m = m e × n = 9 .1×10 -31 ×1 .85×10 12 = 1 .706×10 -18 kg A negligible amount of mass is transferred from wool to polythene. 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Q.14 (a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10−7 C? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Ans.

( a ) Here, charge on sphere A, q A =6 .5×10 -7 C Charge on sphere B, q B =6 .5×10 -7 C Distance between spheres, r=50 cm =0.5 m Force of electrostaticrepulsion between the two spheresisgivenas: F= q A q B 4π ε 0 r 2 Here, ε 0 =Permittivityoffree space= 9×1 0 9 Nm 2 C 2 F= 9×1 0 9 × ( 6.5×1 0 7 ) 2 ( 0.5 ) 2 = 1 .52 × 10 -2 N Theforce between the two spheres is 1.52×1 0 2 N. ( b ) After doubling the amountofcharge, Charge on sphere A, q A = Charge on sphere B, q B =2×6.5×1 0 7 C =1.3×1 0 6 C Asdistance between the spheres is halved, r= 0.5 2 =0.25m Force of repulsion between the two spheresisgivenas, F= q A q B 4π ε 0 r 2 = 9×10 9 ×1 .3×10 -6 ×1 .3×10 -6 ( 0.25 ) 2 = 16×1 .52×10 -2 = 0.243 N Force between the two spheres is 0.243 N. 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Q.15 Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

Ans.

Here,initial charge on each sphere, q = 6 .5 × 10 -7 C Distance between the spheres A and B, r = 0.5 m When sphere A is touched with an uncharged sphere C, theirchargesareequallyshared. ChargeleftonsphereA, q 1 = 6 .5 × 10 -7 2 =3.25×1 0 7 C When sphere C with charge is brought in contact with sphere B with charge q,theirchargesareshared equally. ChargeleftonsphereB, q 2 = ( 6.5+3.25 ) ×10 -7 2 C =4.875×1 0 7 C Asforceofrepulsionisgivenbytherelation, F= 1 4π ε 0 q 1 q 2 r 2 Here, 1 4π ε 0 =9× 10 9 N m 2 C 2 F= 9×10 9 ×3 .25×10 -7 ×4 .875×10 -7 ( 0.5 ) 2 =5 .7×10 -3 N The force of attraction between the two spheres is 5.703×1 0 3 N. 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Q.16 Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field.
Give the signs of the three charges. Which particle has the highest charge to mass ratio?


Ans.

As opposite charges attract each other and same charges repel each other. Therefore, particle 1 and particle 2 are negatively charged and particle 3 is positively charged.
The displacement or the amount of deflection for a given velocity is directly proportional to the charge to mass ratio (emf). As the deflection of particle 3 is the maximum, therefore it has the highest charge to mass ratio.

Q.17

Consider a uniform electric field E = 3×10 3 i ^ NC -1 . (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the samesquare if the normal to its plane makes a 60° angle with the x-axis?

Ans.

( a ) Here,electric field, E = 3×10 3 i ^ NC -1 Magnitude of electric field, | E | = 3×10 3 NC -1 Side of square, s = 10 cm = 0.1 m Area of square, A = s 2 = 0 .01 m 2 Sincetheplane of the square is parallel to the y-z plane Angle between the unitvector normal to the plane and electric field, θ = 0°.

Q.18 What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Ans.

Since all the faces of a cube are parallel to the coordinate axes, therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. Therefore, net flux through the cube is zero.

Q.19 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Determine, B0, ω, k, and λ. (b) Find expressions for E and B.

Ans.

(a) (i) As, c = E0B0B0 = E0c B0 = 120 NC-1 3×108 ms-1 = 400 nT(ii) Angular frequency is given by ω = 2πνω = 3.14×50×106 rad/s = 3.14×108 rad/s(iii)Propagation constant, k = 2πλk = 2×3.146 m = 1.05 rad/m(iv) λ = cνλ = 3×108 ms-150×106 s-1 = 6 m

(b) Suppose a plane electromagnetic wave is propagating in X direction. Then, the electric field vector will be in Y-direction and magnetic field will be in Z-direction. This is because all three vectors are mutually perpendicular to each other.

Electric field vector, E = E0sin(kx – ωt)j^E = (120  NC-1) sin ((1.05  radm-1)x – (3.14×108 rads-1) t)j^ andMagnetic field vector, B = B0sin(kx – ωt)k^B = (400×10-7 T) sin ((1.05  radm-1)x – (3.14×108 rads-1) t)k^  with usual units.

Q.20 Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 103 N m2C-1.
(a) What is the net charge inside the box?
(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

Ans.

( a ) Here,net outward flux through the surface of the box, Ï•=8.0×1 0 3 N m 2 C -1 For a body containing net charge q, flux is givenas: Ï•= q ε 0 Here, ε 0 =Permittivity of free space =8.854×1 0 12 N 1 C 2 m 2 q= ε 0 Ï• =8.854×1 0 12 ×8.0×1 0 3 =7.08×1 0 8 =0.07 μC Net charge inside the box is 0.07 μC. ( b ) If net flux throughthesurfaceis zero, then as per gauss theorem it can be concluded that net charge inside the body is zero. Insuchsituation,the body may have equal amount of positive and negative charges. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@EB4E@

Q.21 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula

E=hν MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaGGGcGaaeyraiabg2da9iaabIgacqaH9oGBaaa@3E9E@

(for energy of a quantum of radiation : photon) and obtain the photon energy in units of eV for different parts of electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

Ans.

As per the relation for photon energyE = hνjoulesE = hν1.6 × 10-19eVAs,ν = cλthenwe get,E = h cλ e-For,wavelength = λ, we obtainEλ=6.63×10-34 Js ( 3×108 ms1)λ ×1.6×10-19 C Similarly wavelengths of different parts of electromagnetic spectrum can be obtained asEλ=1.24 × 10-6λeV a For radio waves let λ = 1 km, thenEλ = 1.24 × 10-61000 meV = 1.24×10-9eVbFor microwaves let λ = 1 cm, thenEλ=1.24 × 10-610-2eV =1.24×10-4eV cFor visible light let λ = 1 µm, thenEλ=1.24 × 10-610-6eV = 1.24 eV .d For X rays let λ = 1nm = 10-9 m,thenEλ = 1.24 × 10-610-9eV = 1.24 ×103eV e For gamma rays let λ = 10-12 m,then Eλ = 1.24 × 10-610-12eV =1.24 ×106eV 

Q.22 A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

Ans.

The square of side 10 cm can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem, total electric flux through all the six faces of the cube is given as:

Ï• total = q ε 0 Electric flux through the square,Ï•= Ï• total 6 = 1 6 q ε 0 Here, ε 0 =Permittivity of free space =8.854×1 0 12 N 1 C 2 m 2 Pointcharge,q=10 μC =10×1 0 6 C Ï• E = 1 6 × 10×1 0 6 8.854×1 0 12 = 1.88×1 0 5 Nm 2 C 1 Electric flux through the square is 1.88×1 0 5 Nm 2 C 1 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1731@

Q.23 A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Ans.

Here,Net charge inside the cube,q=2.0 μC =2×1 0 6 C Net electric flux ( Ï• net ) through the cubic surface is given bytherelation, Ï• net = q ε 0 Here, ε 0 =Permittivity of free space =8.854×1 0 12 N 1 C 2 m 2 Ï• net = 2×1 0 6 8.854×1 0 12 =2.26×1 0 5 Nm 2 C 1 Net electric flux through the surface is 2.26 ×1 0 5 Nm 2 C 1 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@3D3D@

Q.24 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2 × 1010 Hz and amplitude 48 Vm-1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field. (c = 3 x 108 ms-1).

Ans.

aAs,wavelengthofthe wave,λ =cν λ = cν = 3 × 108 ms1 2 ×1010 s1 = 1.5 ×10-2 m bUsing c = E0B0,we get amplitudeoftheoscillatingmagneticfield,B0 = E0cThus B0 = 48 Vm13×108 ms1 = 1.6 ×10-7 T cIn vacuum, the average energy density of electric field (U E),U E = 12 ε0 E2whereas,the average energy density of magnetic field (U B),U B = 12B2µ0ThusU EU B = µ0 ε0 E2 B2= µ0 ε0µ0 ε0 = 1 as, EB=1µ0 ε0 Hence U E = U B

Q.25 A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 NC-1 and points radially inward, what is the net charge on the sphere?

Ans.

Here,radiusofthesphere,r=10cm Electric fieldintensity,E=1.5×1 0 3 NC -1 ( Negativesignforinwardfield ) Distance from the centreofthesphere,d=20 cm =0.2 m Letnetchargeonthesphere=q Electric field intensityis given by the relation, ∵d > r E= q 4π ε 0 d 2 Here, ε 0 =Permittivity of free space =9×1 0 9 Nm 2 C 2 q=E×4π ε 0 d 2 = -1.5×1 0 3 × ( 0.2 ) 2 9×1 0 9 =6.67×1 0 9 C = 6.67 nC Net charge on the sphere is -6.67 nC. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AE0A@

Q.26 Suppose that electric field part of an electromagnetic wave in vacuum is

E = 3.1  N C-1cos (1.8  radm-1) y + (5.4 × 108 rads-1) ti^

(a) What is the direction of propagation?
(b) What is the wavelength λ?
(c) What is the frequency

υ

?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave.

Ans.

aFrom the given equation, it is clear that the direction of propagation of the wave is along negative y-direction i.e –j, because coefficient of y is positive.bComparingthegivenequationwiththeequation,E = E0 cos (k y+ω t),wegetthevalueofk = 1.8  radm-1,ω = 5.4 × 108 rad s-1,E0 = 3.1 NC-1thenwavelengthisgivenbyλ = 2 πk = 2 π1.8= 3.5 m cAs,ν = ω2 π= 5.4×1062 π= 0.86 MHz dc = E0B0thus,B0 = E0c=3.13×108=1.03×10-8=10.3  nT eUsing B = B0 cos k y+ ω t,we get B = 10.03 nT cos (1.8 rad m-1) y+(5.4×108 rad s-1) t k^

Q.27 A point charge causes an electric flux of −1.0 × 103 Nm2C-1 to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?

Ans.

(a) Here,electric flux, Ï•=1.0×103 N m2C-1Radius of theGaussian surface,R=10.0 cmElectric flux passing through a surface depends on the net charge enclosed inside a bodyandit does not depend on the size of the body. If the radiusof theGaussian surface is doubled, thenthe flux passing through the surface remains thesame.(b) Electric flux is given as:Ï•=qε0Here,q = Net charge enclosed by the surfaceε0 = Permittivity of free space = 8.854 × 1012N1C2m2q=Ï•ε0=1.0×103×8.854×1012= 8.854×109 C= 8.854 nCThe value of the point charge is 8.854 nC.

Q.28 A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μCm-2.
(a) Find the charge on the sphere.
(b) What is the total electric flux leaving the surface of the sphere?

Ans.

( a ) Here,diameter of sphere, d = 2.4 m Radius of sphere, r= d 2 = 1.2 m Surface charge densityofsphere, σ= 80.0 μ Cm -2 = 80×1 0 6 Cm -2 Total charge on the surface of the sphereisgivenas: Q =σ×4π r 2 =80×1 0 6 ×4×3.14× ( 1.2 ) 2 =1.447×1 0 3 C Charge on the sphere is 1.447×1 0 3 C. ( b ) Here,Q=1.447×1 0 3 C Total electric flux leaving the surface of a sphere is given by the relation, Ï• total = Q ε 0 Here, ε 0 =Permittivity of free space =8.854×1 0 12 N 1 C 2 m 2 Ï• total = 1.447×1 0 3 8.854×1 0 12 =1.63×1 0 8 NC 1 m 2 The total electric flux leaving the surface of the sphere is 1.63 × 1 0 8 NC 1 m 2 . 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Q.29 An infinite line charge produces a field of 9 × 104 NC-1 at a distance of 2 cm. Calculate the linear charge density.

Ans.

Here,Electric field,E=9 × 104 NC-1Distance,d=2cm=0.02mElectric field produced by the infinite line charges at a distance d having linear charge density λ is given as:E=λ2πε0dλ=2πε0d×EHere,ε0=Permittivity of free space14πε0=9×109 Nm2C2λ=0.02×9 × 1042×9×109= 107 Cm-1The linear charge density is 0.1 μCm-1.

Q.30 About 5 % of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) at a distance of 1 m from the bulb.
(b) at a distance of 10 m.
Assume that the radiation is emitted isotropically and neglect reflection.

Ans.

Power rating of the bulb, P = 100 WIt is given that 5 % of its power is convered into visible radiation.Power of visible radiation, P’ = 5100 × 100 = 5 WaAs average intensity of radiation at 1 mdistance from the bulb, I = Power4 π r2Thus, intensity, I = 5 W4×π×1 m2 = 0.4 Wm2.bAverage intensity of radiation at 10 mDistance from the bulb,I’ = 5 W4×π×102 m2 = 0.004Wm-2

Q.31

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0×10-22Cm-2. What is E:(a) in theouter region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

Ans.


The situation of the problem is represented in the given figure.

Here, A and B are two parallel plates close to each other. Outer region of plate A is labeled as I, the region between the plates, A and B, is labelled as II and outer region of plate B is labelled as III.

Here,surfacecharge density, σ = 17 .0×10 -22 Cm -2 Sincecharge is notenclosed by the respective platesin the regions, I and III, Electric field, E = 0,inregionsI and III. Thus, ( a )and( b ) have net electric field zero. ( c )Electric field E in region II is given by, E = σ ε 0 Here, ε 0 =8 .854×10 -12 N -1 C 2 m -2 E = 17 .0×10 -22 Cm -2 8 .854×10 -12 N -1 C 2 m -2 = 1 .92×10 -10 NC -1 Electric field between the plates is 1 .92×10 -10 NC -1 .

Q.32 An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 NC−1 in Millikan’s oil drop experiment. The density of the oil is 1.26 gcm−3. Estimate the radius of the drop. (g = 9.81 ms−2; e = 1.60 × 10−19 C).

Ans.

Here, excess electrons on an oil drop, n = 12Magnitude of Electric field, E = 2.55×104 NC-1Density of oil, ρ = 1.26 gmcm-3 = 1.26 × 103 kgm-3Acceleration due to gravity, g = 9.81 m s-2Charge on an electron, e = 1.6×10-19 CLet radius of oil drop = rAs the oil drop is stationary,Force due to electric field = Weight of the oil dropF = WAs F=Eq and W=mgEq = mg(i) Charge on oil drop, q= ne→(ii)Mass of oil drop, m= Volume of the oil drop × Density of oil=43πr3×ρ(iii)Substitutingequations (ii) and (iii),inequation(i),wegetEne=43πr3×ρgr=3Ene4πρg3=3×2.55×104×12×1.6×10-194×3.14×1.26×103×9.813=946.09×10-213=9.82×10-7 m= 9.82×10-4 mmThe radius of the oil drop is 9.82 × 104 mm.

Q.33 Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?


Ans.

(a) As the field lines must be normal to the surface of the conductor, therefore, the field lines showed in figure (a) do not represent electrostatic field lines.

(b) As the electrostatic field lines do not start from a negative charge, therefore, the field lines showed in figure (b) do not represent electrostatic field lines.

(c) Since the field lines emerge from the positive charges and repel each other, therefore, the field lines showed in figure (c) represent electrostatic field lines.

(d) As two electrostatic field lines cannot intersect each other, therefore, the field lines showed in figure (d) do not represent electrostatic field lines.

(e) The field lines showed in figure (e) do not represent electrostatic field lines because electrostatic field lines never form closed loops.

Q.34 Use the formula

λmT

= 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?

Ans.

Accordingto Wein’slaw,λmT = 0.29 cm K, thusT = 0.29 cm K λmwhich implies that, T = 0.29 K forλm = 1 cm .a Forλm = 10-10cmwe get, T = 0.29 cm K10-10 cm= 2.9×109 K b Forλm=10-6 m=10-4 cmwe get,T = 0.29 cmK10-4 cm= 2900  KSimilarly, temperatures for other wavelengths can be found. These numbers tell us the temperature ranges required for obtaining radiations in different parts of the electromagnetic spectrum. Thus to obtain visible radiation of wavelength 5 x 10-7m, the source should have a temperature of about T = 0.29 cmK5×10-5 cm= 5800 KT≃6000 K.

Q.35 In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 105 NC−1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10−7 C-m in the negative z-direction?

Ans.

Here, dipole moment of the system, p = q × dl = −10−7 Cm

The rate of increase of electric field per unit length ,

dE dl = 10 5 NC -1 Force experienced by the system is given as: F = qE =q dE dl ×dl =p× dE dl =-10 -7 ×10 5 = -10 -2 N MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8948@ The forceonthedipole is1 0 2 N in the negative z direction Angle between electric field and dipole moment,θ =180°. Torque is given as:τ =pE sinθ= pE sin180°= 0 Torque experienced by the system is zero. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@EC3C@

Q.36 (a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.
(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig.1.36(b)].
(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.


Ans.

(a) Consider a Gaussian surface lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero.

Let charge inside the conductor=q  According to Gausss law,wehaveFlux,Ï•=E.ds=qε0Here, ε0=Permittivity of free spaceAs net electric field inside a charged conductor, E=0qε0=0Asε00q=0Chargeinsidetheconductor=0.EntirechargeQontheconductorappearsontheoutersurfaceoftheconductor.

(b) Another conductor B having charge +q is inserted inside conductor A and it is insulated from A. Therefore, a charge of amount −q will be induced in the inner surface of conductor A and +q is induced on the outer surface of conductor A. Since the outer surface of conductor A has originally a charge of amount Q, therefore, total charge on the outer surface of conductor A becomes Q + q.

(c) A sensitive instrument can be shielded from the strong electrostatic field in its environment by enclosing it in a metallic cover.

Q.37 A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is, where is the unit vector in the outward normal direction, and is the surface charge density near the hole.

Ans.

Consider a conductor with a tinyholeintoitssurface. Electric field inside the hole =0 Suppose Eiselectric field outside the conductor,qis theelectricchargeandσisthesurfacechargedensity. Electriccharge,q=σ×ds According to Gausss law,wehave ∮ E . ds = q ε 0 Here, ε 0 =Permittivityoffreespace Eds= σ×ds ε 0 E= σ ε 0 or E = σ ε 0 n ^ Electric field just outside the conductor is σ ε 0 n ^ . This field is superposition of field due to filleduphole and the field due to the rest of thecharged conductor. These twofields are equal and opposite inside the conductor.Therefore,thereisnoelectricfieldinside theconductor.However,thesefieldsareequal in magnitude and direction outside the conductor. The electricfieldatP, due to eachpart= 1 2 E = σ 2 ε 0 n ^ 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Q.38 Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs:

(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space)

(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen)

(c) 2.7K ( temperature associated with the isotropic radiation filling all space – thought to be a relic of the big-bang origin of the universe)

(d) 5890 A – 5896 A (double lines of sodium)

(e) 14.4 keV (energy of a particular transition in Fe nucleus associated with a famous high resolution spectroscopic method.

Ans.

aShort radio wave as wave length is of the order of 10-2 m.bShort radio wave as frequency is of the order of 10-9 Hz.c AsλmT = 0.29 cm K, thus,λm = 0.29T = 0.29 cm K2.7 K= 0.11 cmThus wavelength correspondsto microwaveregionofe.m.spectrum .d Visible radiations Yellow light as given wavelengths are of the order of 10-7m.e E = 14.4 keV = 14.4 × 103 × 1.6 × 10-19 JUsing ,E = hνν = Eh= 14.4 × 103 × 1.6 x 10-19 J6.63 ×10-34Js= 3.47 × 1018 Hz .This corresponds to X rays.

Q.39 Answer the following questions:
(a) Long distance radio broadcasts use short wave bands. Why?
(b) It is necessary to use satellites for long distance TV transmission. Why?
(c) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting earth. Why?
(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?
(e) If the earth did not have its atmosphere, would its average surface temperature be higher or lower than what it is now?
(f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life of earth. What might be the basis of this prediction?

Ans.

(a) As long distance radio broadcasts make use of sky waves where Ionosphere of earth’s atmosphere reflects the radiations of this range.

(b) For every long distance transmission of TV signals, a very high frequency is required. Waves of this frequency just pass through the ionosphere, without being reflected, so a satellite is required to return the signals to the earth.

(c) As X-rays has smaller wavelength, it can be absorbed by the earth. Hence X ray astronomy is possible only from satellites orbiting the earth but visible waves and radio waves can pass through the atmosphere therefore we can work with visible waves and radio waves on earth’s surface.

(d) The small ozone layer on top of the stratosphere is crucial for human survival as it absorbs harmful radiations (ultraviolet radiations) present in the sunlight and prevents it from reaching the surface of the earth. Ultraviolet radiations are harmful for the life on earth.

(e) In this case, there will be no green house effect. So the earth will be at lower temperature.

(f) In the case of worldwide nuclear war the sky may get overcast with clouds. These clouds will prevent sunlight from reaching many parts of the globe. Thus earth will be as cool as in winter.

Q.40 Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Ans.

Consider a long thin wire AB (as shown in the figure) of uniform linear charge density λ.

Take a point S at a perpendicular distance l from the mid-point O of the wire, as shown in the given figure.

Let E be the electric field at point S because of wire AB. Consider a small element of length dx on the wire with OZ = x Let q be the charge on this element. q=λdx Electric field due to the element, dE= 1 4π ε 0 λdx ( SZ ) 2 As,SZ= l 2 + x 2 dE= λdx 4π ε 0 ( l 2 + x 2 ) Theelectricfieldcanberesolvedintotworectangular components:dEcosθperpendiculartoABanddEsinθ paralleltoAB. Only the perpendicular component dEcosθaffects point S. Effective componentofelectric field at point S due to the element dx, dE 1 =dEcosθ dE 1 = λdxcosθ 4π ε 0 ( l 2 + x 2 ) (i) InΔSZO, tanθ= x l x=ltanθ(ii) Differentiating equation ( ii ), we get dx dθ =l sec 2 θ dx=l sec 2 θdθ(iii) From equation ( ii ), x 2 + l 2 = l 2 + l 2 tan 2 θ l 2 ( 1+ tan 2 θ )= l 2 sec 2 θ x 2 + l 2 = l 2 sec 2 θ(iv) Substituting equations ( iii ) and ( iv ) in equation ( i ), we get d E 1 = λl sec 2 θdθ 4π ε 0 l 2 sec 2 θ ×cosθ d E 1 = λcosθdθ 4π ε 0 l (v) The wire is so long that angleθtends from – π 2 to+ π 2 . Integrating equation ( 5 ), we obtain the value of E 1 as, π 2 π 2 d E 1 = π 2 π 2 λcosθdθ 4π ε 0 l E 1 = λ 4π ε 0 l [ sinθ ] π 2 π 2 = λ 4π ε 0 l ×2 E 1 = λ 2π ε 0 l Theelectric field due to long wire is λ 2π ε 0 l . 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Q.41 It is now established that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge (+2/3) e, and the ‘down’ quark (denoted by d) of charge (−1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

Ans.

Here, total number of quarks in a proton=3Let number of up quarks in a proton=n Number of down quarks in a proton = 3-nTotal charge on the proton is given as:(23e)n+(13e)(3-n)=+e23ene+ne3=ene(23+13)=2ene=2en=2Numberofupquarksinaproton=2Numberofdownquarksinaproton=32=1Aprotoncanberepresentedas:pUUdHere, total number of quarks in a proton=3Let number of up quarks in a neutron=n Number of down quarks in a neutron = 3-nTotal charge on the neutron is given as:(23e)n+(13e)(3-n)=023ene+ne3=one(23+13)=0ne=en=1Numberofupquarksinaneutron=1Numberofdownquarksinaneutron=31=2Aneutroncanberepresentedas:pUdd

Q.42 (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

Ans.

(a) Suppose the equilibrium of the test charge is stable. When a test charge is displaced from its equilibrium position in any direction, it experiences a restoring force towards a null point, where the electric field is zero. It shows that there is a net inward flux of electric field through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface not enclosing any charge is zero. Therefore, the equilibrium of the test charge cannot be stable.
(b) The mid-point of the line joining the two like charges is the null point. When a test charged is slightly displaced along the line, it experiences a restoring force. However, if it is displaced normal to the joining line, then the net force takes it further away from the null point. Therefore, the equilibrium cannot be stable.

Q.43 A particle of mass m and charge (−q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL2/ (2m vx2).
Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.

Ans.

Here,Charge on the particle of mass m= qVelocity of particle=vxLength of plates=LMagnitude of the uniform electric field=EMechanical forceisgivenas: F=Mass m×Acceleration aa=FmElectricforceisgivenas:F=qEAcceleration,a=qEm(i)Time taken by the particle to cross the field is given as:t=distancevelocity=lengthofplatevelocityofparticle=Lvx(ii)Along vertical direction,initial velocity,u=0According to third equation of motion,the vertical deflection s of the particle can beobtained as,s=ut+12at2s=0+12qEmLvx2s=qEL22mvx2(iii)Vertical deflection of the particle at the far edge of the plate isqEL22mvx2. This is exaclysimilar to the motion of horizontal projectiles under theeffectofgravity.

Q.44 Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx= 2.0 × 106 m s−1. If E between the plates separated by 0.5 cm is 9.1 × 102 NC-1, where will the electron strike the upper plate? (| e | =1.6 × 10−19 C, me = 9.1 × 10−31 kg.)

Ans.

Here, velocity of particle, v x = 2 .0×10 6 ms -1 Separation of the plates, d = 0.5 cm = 0.005 m Electric field between the plates, E=9 .1×10 2 NC -1 Charge of electron, q = 1 .6 × 10 -19 C Mass of electron, m e = 9 .1 × 10 -31 kg Let the electron strikes the upper plate at the end of plate L,assoonasits deflection is d.d= qEL 2 2m v x 2 L= 2×0.005×9.1× 10 31 × ( 2.0×1 0 6 ) 2 1.6 × 1 0 19 ×9.1×1 0 2 = 0.025× 10 2 = 2.5× 10 4 =1.6× 10 2 m =1.6cm Theelectron will strike the upper plate after travelling 1.6 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D963@

Q.45 Two charges 5 × 10−8 C and −3 × 10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Ans.

Here,q1=5×108Cq2=3×108CDistance between the two charges, d=16 cm=0.16 mLet potential be zero at a distance s cm from charge q1=5×108Cr1=s×102m;r2=(16s)×102mLet electric potential at distance s=0Electricpotentialatapointisthesumofpotentialscausedbychargesq1andq2respectively.V=q14πε0r+q24πε0(dr)(i)Where,ε0= Permittivity of free spaceFor V=0, equation (i) becomes0=q14πε0r+q24πε0(dr)q14πε0r=q24πε0(dr)q1r=q2(dr)5×108r=(3×108)(0.16r)0.16r0.16r=85r=0.1m=10cmThe potential is zero at a distance of 10 cm from the positive charge q1If the point lies outside the charges,{14πε°}[5×108r  3×108(r0.16)]=0r=40cm r = 40 cm from positive charge towards negative charge on extended line.

Q.46 A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.

Ans.

Consider a regular hexagonABCDEF, havinganequalamountofchargeq ateachofitsvertices. Here, charge, q = 5 μC = 5×10 -6 C Side of hexagon, l = 10 cm Distance of each vertex from thecentre O, d = 10 cm = 0.1 m Aselectricpotentialisscalar, Electric potential at Oisgivenas V = 6q 4π ε 0 d Here, ε 0 = Permittivity of free space 1 4π ε 0 = 9×10 9 NC -2 m -2 V = 6×9×10 9 NC -2 m -2 ×5×10 -6 C 0.1 m = 2 .7×10 6 V The potential at the centre of the hexagon is 2 .7×10 6 V.

Q.47 Two charges 2 μC and −2 μC are placed at points A and B 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?

Ans.

(a) The plane normal to AB and passing through its middle point has zero potential everywhere and it represents an equipotential surface of the given system.
(b) The direction of the electric field at every point on this surface is along normal to the plane in the direction of AB.

Q.48 A spherical conductor of radius 12 cm has a charge of 1.6 × 10−7 C distributed uniformly on its surface. What is the electric field
(a) Inside the sphere
(b) Just outside the sphere
(c) At a point 18 cm from the centre of the sphere?

Ans.

(a)Here, Radius of the spherical conductor, r=12 cm =0.12 mCharge over the spherical conductor, q=1.6×107 CIf there is field inside the conductor, then charges will move to neutralize it.Electric field inside the spherical conductor =0 (b) Electric field just outside the spherical conductor isgiven as:E=q4πε0r2Here,ε0= Permittivity of free space14πε0=9×109Nm2C2E=9×109×1.6×107(0.12)2=105NC1The electric field just outside the sphere is 105NC1. (c) Let Electric field at a point 18 m from the centre of the sphere=E1Distance of the point from the centre of sphere, d =18 cm=0.18 m E1=q4πε0r2=9×109×1.6×107(0.18)2=4.4×104NC1The electric field at a point 18 cm from the centre of the sphere is 4.4×104NC1.

Q.49 A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Ans.

Asinitiallyairwasfilledbetwentheplatesofparallelplatecapacitor,Dielectricconstantofair,K=1Initialcapacitanceoftheparallelplatecapacitor=8pFLetareaofeachplateofcapacitor=ALetinitialdistancebetweentheplatesoftheparallelplatecapacitor=dCapacitanceoftheparallelplatecapacitorisgivenas:C1=ε0Ad=8pF(i)Whendistancebetweentheplatesisreducedtohalf,thennewdistance,d=d2Dielectricconstantofsubstancefilledbetweentheplates,K=6Capacitanceoftheparallelplatecapacitorbecomes,C2=Kε0Ad=6ε0Ad2=6×2×ε0Ad=12×ε0Ad=12×8[Usingequation(i)]=96pFThecapacitancebetweentheplatesis96pF.

Q.50 Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

Ans.

(a) Capacitance of each capacitor, C=9 pFSince the three capacitors are connected in series,Equivalent capacitance of the combination of capacitors is given by the relation,1Cs=1C1+1C2+1C31Cs=19+19+19=39=13Cs=3pFTotal capacitance of the combination is 3pF. (b) Here, supply voltage, V=120 VSince the three capacitors are connected in series,Thepotential difference (V) across each capacitor is equal to onethird of the supply voltage.V=V3=1203=40VThe potential difference across each capacitor is 40 V.

Q.51 Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Ans.

(a)Given,C1=2pFC2=3pFC3=4pFFor the parallel combination of the capacitors, the total capacitance of the combination, Cp=C1+C2+C3=2+3+4=9pFTotal capacitance of the combination is 9 pF.(b) Here, supply voltage, V=100 VThe voltage through each of the three capacitors is same, V=100 VCharge on the capacitor of capacitance C and potential difference V is given by the relation,q=VC (i)For C=2 pF,Charge,q1=VC=100×2=200 pF=2×1010CFor C=3 pF,Charge,q2=VC=100×3=300 pF=3×1010CFor C=4 pF,Charge,q3=VC=100×2=200 pF=2×1010C

Q.52 In a parallel plate capacitor with air between the plates, each plate has an area of 6 ×10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

Ans.

Here, area of each plate of the parallel plate capacitor, A=6×103 m2Separation between the plates, d=3 mm =3×103 mSupply voltage, V=100 VCapacitance of a parallel plate capacitor is given by the relation,C=ε0AdHere,ε0=Permittivity of free space=8.854×1012 N1m2C2C=8.854×1012×6×1033×103=17.71×1012F=17.71pFPotentialisrelatedtothechargeqandcapacitanceCas:V=qCq=VC=100×17.71×1012=1.771×109CCapacitance of the capacitor is 17.71 pF and charge on each plate is 1.771 ×109 C.

Q.53 Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
(a) While the voltage supply remained connected.
(b) After the supply was disconnected.

Ans.

( a ) Whilethevoltagesupplyremainedconnected, potential across the plates remains 100 V. Here,dielectric constant of the mica sheet, k=6 Initial capacitance, C=1.771 × 1 0 11 F Newcapacitanceisgivenby, C’=kC C’=6×1.771×1 0 11 =106pF Supply voltage, V=100 V Newchargeisgivenby, q’=C’V =106×1 0 12 ×100 =1.06× 1 0 8 ( b ) Afterthesupplywasdisconnected,theamountof charge remainsconstant. Here,dielectric constant, k=6 Initial capacitance, C=1.771×1 0 11 F Newcapacitanceisgivenas:C’=kC =6×1.771×1 0 11 =106pF Charge,q=1.771×1 0 9 C Potential across the plates is given bytherelation: V’= q C = 1.771×1 0 9 106×1 0 12 =16.7V MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@708D@

Q.54 A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Ans.

Here,capacitor of the capacitance, C=12 pF = 12×1 0 12 F Potential difference, V=50 V Electrostatic energy stored in the capacitor is given as: E= 1 2 C V 2 = 1 2 ×12×1 0 12 × ( 50 ) 2 =1.5× 10 8 J The electrostatic energy stored in the capacitor is 1.5× 10 8 J. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@0B20@

Q.55 A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Ans.

Here,capacitance of the capacitor, C=600 pF Supplyvoltage, V=200 V Electrostatic energy stored in the capacitor is given by therelation, E= 1 2 C V 2 = ( 600× 10 12 )× ( 200 ) 2 2 =1.2× 10 5 J Whensupply voltageis disconnected from the capacitor and another capacitor of capacitance C= 600 pF is connected to it,then equivalent capacitance ( C ) of the combination is givenas: 1 C = 1 C + 1 C = 1 600 + 1 600 = 1 300 C=300pF New electrostatic energy can be obtained as: E’= 1 2 ×C× V 2 = 1 2 ×300× ( 200 ) 2 =0.6× 10 5 J Lossofelectrostatic energy=EE =1.2× 10 5 0.6× 10 5 =0.6× 10 5 =6× 10 6 J The electrostatic energy lost in the process is 6× 10 6 J. 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Q.56 A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10−9 C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

Ans.

Here,charge located at origin, q=8 mC =8×1 0 3 C Magnitude of the charge, which is tobecarried from point P to point Q, q 1 =2×1 0 9 C All the points are shown in the given figure. Distanceofpoint P, d 1 =3 cm, from the origin along zaxis. Distanceofpoint Q, d 2 =4 cm, from the origin along yaxis. Potential at point Pisgivenas: V 1 = q 4π ε 0 × d 1 Potential at point Qisgivenas: V 2 = q 4π ε 0 × d 2 Sincework done ( W ) by the electrostatic force is independent of the path, W= q 1 [ V 2 V 1 ] = q 1 [ q 4π ε 0 × d 2 q 4π ε 0 × d 1 ] = q 1 q 4π ε 0 [ 1 d 2 1 d 1 ](i) Here, 1 4π ε 0 =9× 10 9 N m 2 C 2 W=9× 10 9 ×8×1 0 3 ×( 2×1 0 9 )[ 1 0.04 1 0.03 ] =144×1 0 3 ×( 25 3 ) =1.27J Work done during the process is 1.27 J. 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Q.57 A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Ans.

Here, side of the cube=b Charge at eachvertexofthecube=q A cube of side b is shown in the given figure. Here,d=Diagonalofthecubeofeachsideb d= b 2 + b 2 = 2 b 2 = 2 b d=b 2 l=Length of the diagonal of the cube l= d 2 + b 2 = ( 2 b ) 2 + b 2 = 2 b 2 + b 2 = 3 b 2 = 3 b Distancebetweenthecentreofcubeandeach vertex,r= l 2 = 3 b 2 Electric potential ( V ) at the centre of the cube is becauseof the presence of eightcharges at the vertices. V= 8q 4π ε 0 r = 8q 4π ε 0 ( 3 b 2 ) = 4q 3 π ε 0 b The potential at the centre of the cube is 4q 3 π ε 0 b . The electric field at the centre of the cube, due to the eight charges, gets cancelled, because the charges are distributed symmetrically w.r.t. the centre of thecube. The electric field is zero at the centre. 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Q.58 Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:
(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Ans.

Two charges placed at points A and B are represented in the given figure. O is the midpointof the line joining the two charges. Amount of charge located at A, q 1 = 1.5μC Amount of charge located at B, q 2 = 2.5 μC Distance between the charges, d = 30 cm= 0.3 m ( a ) Let V 1 and E 1 be the electric potential and electric field respectively at O. V 1 =Potential due to charge at A+Potential due to charge at B V 1 = q 1 4π ε 0 ( d 2 ) + q 2 4π ε 0 ( d 2 ) = 1 4π ε 0 ( d 2 ) ( q 1 + q 2 ) Here, ε 0 =Permittivity of free space 1 4π ε 0 =9× 10 9 N C 2 m 2 V 1 = 9× 10 9 × 10 6 ( 0.30 2 ) ( 2.5+1.5 ) =2.4× 10 5 V E 1 =Electric field due to charge q 2 Electric field due to charge q 1 = q 2 4π ε 0 ( d 2 ) 2 q 1 4π ε 0 ( d 2 ) 2 = 9× 10 9 ( 0.30 2 ) 2 × 10 6 ×( 2.51.5 ) =4× 10 5 V m 1 At themidpoint,the potential is 2.4 × 1 0 5 V and electric field is4× 1 0 5 V m 1 . The directionoftheelectricfield is from the larger charge to the smaller charge. 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( b ) Consider a point Z such that thenormal distance OZ=10 cm=0.1 m, as shown in thegiven figure. V 2 and E 2 respectivelyarethe electric potential and electric field at Z. From the figure,it can be observed that distance, BZ=AZ = ( 0.1 ) 2 + ( 0.15 ) 2 =0.18m V 2 = Electric potential due to A + Electric Potential due to B = q 1 4π ε 0 ( AZ ) + q 1 4π ε 0 ( BZ ) = 9× 10 9 × 10 6 0.18 ( 1.5+2.5 ) =2× 10 5 V Electric field due to q at Zisgivenby, E A = q 1 4π ε 0 ( AZ ) 2 = 9× 10 9 ×1.5× 10 6 ( 0.18 ) 2 =0.416× 10 6 V m 1 E= E A 2 + E B 2 +2 E A E B cos2θ Here, 2θis the angle,AZB From the figure, we get cosθ= 0.10 0.18 = 5 9 =0.5556 θ= cos 1 0.5556 =56.25 2θ= 112.5 o cos2θ=0.38 E= ( 0 .416×10 6 ) 2 × ( 0 .69×10 6 ) 2 +2×0.416×0 .69×10 12 ×( -0.38 ) =6.6× 10 5 V m 1 The potential at a point 10 cm ( perpendicular to the midpoint ) is 2.0×105 Vand electric field is 6.6 ×1 0 5 V m 1 . 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Q.59 A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.
(a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Ans.

(a) The magnitude of charge placed at the centre of the shell is +q. Therefore, a charge −q will be induced to the inner surface of the shell.

Total charge on the inner surface of the shell = q.Surface charge density at the inner surface of the shellis given as:σ1=TotalchargeInnersurfacearea=q4πr12(i)A charge of magnitude +q induces on the outer surface of the shell. Since a charge of magnitude Q is placed on the outer surface of the shell. Total charge on the outer surface of the shell= Q+q Surface charge density at the outer surface of the shell is given as:σ2=TotalchargeOutersurfacearea=Q+q4πr22(i)

(b) Yes, the electric field intensity inside a cavity with no charge is zero, even if the shell is not spherical and has any irregular shape. When a closed loop is taken such that a part of it is inside the cavity along a field line and the rest is inside the conductor; the net work done by the electric field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field inside a cavity with no charge is zero, whatever is the shape.

Q.60

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by ( E 2 E 1 ) n ^ = σ ε 0 Where, ( n ^ ) is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of ( n ^ )is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ n ^ ε 0 . (b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

Ans.

( a ) Lete lectric field on one side of a charged body is E 1 and electric field on the other side of the same body is E 2 . If theinfinite plane charged body has a uniform thickness, then Electric field due to one surface of the charged body is given as: E 1 = σ 2 ε 0 n ^ (i) Here, n ^ = Unit vector normal to the surface at a point σ = Surface charge density at that point Electric field due to the other surface of the charged bodyisgivenas: E 2 = σ 2 ε 0 n ^ (ii) Electric field at any point due to the two surfacesis givenas: E 2 E 1 = σ 2 ε 0 n ^ + σ 2 ε 0 n ^ = σ ε 0 n ^ ( E 2 E 1 ). n ^ = σ ε 0 As, inside the closed conductor, E 1 = 0, E = E 2 = σ 2 ε 0 n ^ Electric field just outside the conductor is σ ε 0 n ^ . ( b )If a charged particle is moved from one point tothe other on a closed loop, Work done by the electrostatic field=0 The tangential component ofelectrostatic field is continuous from one side of a charged surface to the other. 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Q.61 A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Ans.

Charge density of the long charged cylinder of length L and radius a= λ. Another cylinder of same length surrounds the previous cylinder,its radius= b Let electric field produced in the space between the two cylinders=E Accordingto Gausss theorem,electric flux through the cylindericalGaussian surface is given by, Ï•=( 2πr )L Here, r= Distance of a point from the common axis of the twocylinders Let total charge on the cylinder=q AccordingtoGauss’stheorem Ï•=E( 2πrL )= q ε 0 Here,q=Charge on the inner sphere of the outer cylinder Since,q=λL ε 0 =Permittivity of free space E( 2πrL )= λL ε 0 E= λ 2π ε 0 r The electric field in the space between the two cylinders is λ 2π ε 0 r . 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Q.62 In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:
(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?

Ans.

Here,distance between electronandproton in a hydrogen atom,d=0.53 A o Charge ofelectron, q 1 =1.6×1 0 19 C Charge ofproton, q 2 =+1.6×1 0 19 C ( a ) Thepotential at infinity is zero. P.E.ofthesystem=P.E.atinfinityP.E.atdistanced =0 q 1 q 2 4π ε 0 d Here, ε 0 =Permittivity of free space 1 4π ε 0 =9× 10 9 N m 2 C 2 P.E.=0 9× 10 9 × ( 1.6×1 0 19 ) 2 0.53× 10 10 =43.47×1 0 19 J Since1.6×1 0 19 J=1eV P.E.=43.47×1 0 19 J = 43.47×1 0 19 1.6×1 0 19 eV =27.2 eV P.E. of the system is 27.2 eV. ( b ) As,K.E. intheorbit= 1 2 ( P.E. ) K.E.= 1 2 ×( 27.2 ) =13.6eV Total energy=13.627.2 =13.6 eV The minimum work required to free the electron is 13.6 eV. ( c ) When zero of P.E. is taken, d 1 =1.06 A o P.E. of the system=P.E . at d 1 P.E. at d = q 1 q 2 4π ε 0 d 1 27.2eV = 9× 10 9 × ( 1.6×1 0 19 ) 2 1.06× 10 10 27.2eV =21.73×1 0 19 J27.2eV =13.58eV27.2eV =13.6eV 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Q.63

If one of the two electrons of a H 2 molecule is removed, we get a hydrogen molecular ion Å . In the ground state of an H 2 + , the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Åfrom each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Ans.

The system of two protons and one electron is shown in the given figure. Here,charge on theelctron , q 1 =-1.6×1 0 19 C Charge on proton 1, q 2 =1.6×1 0 19 C Charge on proton 2, q 3 =1.6×1 0 19 C Distance between q 1 and q 2 , d 1 =1 ×1 0 10 m Distance between q 2 and q 3 , d 2 =1.5 ×1 0 10 m Distance between q 3 and q 1 , d 3 =1 × 1 0 10 m Taking zeroofpotential energy at infinity P.E. of the system,V= q 1 q 2 4π ε 0 d 1 + q 2 q 3 4π ε 0 d 2 + q 3 q 1 4π ε 0 d 3 Here, 1 4π ε 0 =9× 10 9 N m 2 C 2 V=9× 10 9 [ ( -1.6×1 0 19 )×( 1.6×1 0 19 ) 1×1 0 10 + ( 1.6×1 0 19 ) 2 1.5 ×1 0 10 + ( 1.6×1 0 19 )×( -1.6×1 0 19 ) 1×1 0 10 ] = 9× 10 9 1 0 10 × 10 38 ×[ -1.6×1.6+ ( 1.6 ) 2 1.5 ( 1.6 ) 2 ] =9× 10 19 ×3.42 =30.78× 10 19 J Thepotential energy of the system is 19.2 eV. 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Q.64 Two charged conducting spheres of radii a and b are connected to each other by a wire.
What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Ans.

Let the radius of a sphere A=a Letcharge onthe sphereA= Q A Let capacitance of the sphereA= C A Let radius of a sphere B=b, Let charge onsphereB= Q B Letcapacitance of the sphereB= C B Asthe two spheres are connected with a wire, their potentials ( V ) become equal. Let electric field of sphere A=E A Let electric field of sphere B=E B Ratio,of E A to E B , E A E B = Q A 4π ε 0 × a 2 × b 2 ×4π ε 0 Q B E A E B = Q A Q B × b 2 a 2 (i) But, Q A Q B = C A V C B V and C A C B = a b (ii) Substituting the value of ( ii ) in ( i ), we get E A E B = a b × b 2 a 2 = b a The ratio of electric fields at the surface is b a . 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Q.65 Two charges −q and +q are located at points (0, 0, − a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.
(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

Ans.

( a ) Here,charge q is at point( 0, 0, a ) and charge +qis at point( 0, 0, a ). Theyform anelectric dipole. Point ( 0, 0, z ) is locatedon the axisof this dipole and point ( x, y, 0 ) is normal tothe axis of the dipole. Electrostatic potential at point ( x, y, 0 )=0 Electrostatic potential at point ( 0, 0, z ) is given as: V= 1 4π ε 0 ( q za )+ 1 4π ε 0 ( q z+a ) = q( z+az+a ) 4π ε 0 ( z 2 a 2 ) = 2qa 4π ε 0 ( z 2 a 2 ) = p 4π ε 0 ( z 2 a 2 ) Here, ε 0 = Permittivity of free space p = Dipole moment of the system of two charges = 2qa ( b ) Sincethedistance r is much greater than half of the distancebetween the two charges. The potential ( V ) at a distance r is inversely proportional to square of the distancei.e., V 1 r 2 ( c ) Electrostatic potential at point ( 5, 0, 0 )isgivenas: V 1 = q 4π ε 0 1 ( 7 ) 2 + ( a ) 2 + q 4π ε 0 1 ( 7 ) 2 + ( a ) 2 = q 4π ε 0 49+ a 2 + q 4π ε 0 1 49+ a 2 =0 Workdoneinmovingasmalltestchargefrompoint ( 5, 0, 0 )topoint( 7, 0, 0 )=Charge( V 2 V 1 )=0 No work is done in moving a small test charge from point ( 5, 0, 0 ) to point (7,0, 0) along the x axis. Aswork done by electrostatic field in moving atest charge between two points is independent of the path connecting the two points, Workdonewillcontinuetobezeroalongeverypath. 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Q.66 Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).


Ans.

The system of three charges forms an electric quadrupole.Thegivenfigureshowsaquadrupole.Here,charge +q is located at point X, charge 2q is located at point Y and Charge +q located at point ZLet XY=YZ = aLet YP=rLet PX=r+aLet PZ=raElectrostatic potential caused by the system of three charges at point P is given as:V=14πε0[qXP2qYP+qZP]=14πε0[qr+a2qr+qr-a]=q4πε0[r(r-a)2(r+a)(r-a)+r(r+a)r(r+a)(r-a)]=q4πε0[r2ra2r2+2a2+r2+rar(r2a2)]=q4πε0[2a2r(r2a2)]=2qa24πε0r3(1a2r2)Asra≫1,ar≪1a2r2 is negligibly small.It is clear that, potential,V1r3However, for a dipole V1r2,andforamonopoleV1r

Q.67 An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 kV. A large number of 1 μF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.

Ans.

Here,total required capacitance, C=2 μF Potential difference, V=1 kV=1000 V Capacitance of each capacitor, C 1 =1μF Maximumpotentialdifferenceacrosseach capacitor, V 1 =400 V Suppose a number of capacitors are connected in series and these series circuits areconnected in parallel ( row ) to each other. Since, potential difference across each row=1000 V Potential difference across each capacitor= 400 V. Numberof capacitors in each row= 1000 400 =2.5 There are three capacitors in each row. Capacitance of each rowof3capacitorsof1μFeach= 1 1+1+1 = 1 3 μF Let there are n rows, each having three capacitors, connected in parallel. Equivalent capacitance of circuit= 1 3 + 1 3 + 1 3 +nterms = n 3 But,capacitance of circuit=2μF n 3 =2 n=6 Numberof rows of three capacitors in the circuit=6 Numberofcapacitors required for the given arrangement= 6×3 = 18 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Q.68 What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of 5F or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

Ans.

Here, capacitance of a parallel plate capacitor, V=2 FSeparation between the two plates, d=0.5 cm =0.5×102 mCapacitance of the parallel plate capacitor is given as:C=ε0AdA=Cdε0Here,ε0= Permittivity of free space =8.85×1012 C2N1m2A=2×0.5×1028.85×1012=1130km2The area of the plates is too large. Inordertoavoidthissituation, the capacitance is taken in the range of 5F.

Q.69 Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply, determine the charge and voltage across each capacitor.


Ans.

Here,c apacitance of capacitor C 1 = 100 pF Capacitance of capacitor C 2 = 200 pF Capacitance of capacitor C 3 = 200 pF Capacitance of capacitor C 4 = 100 pF Voltageapplied, V=300 V Thec apacitors C 2 and C 3 are connected in series. Let equivalent capacitance of C 2 and C 3 =C 1 C = 1 200 + 1 200 = 2 200 C=100pF Thec apacitors C 1 and C are in parallel. Let equivalent capacitance of C 1 and C=C” C”=C+ C 1 =100+100 =200pF Thecapacitors C”and C 4 are connected in series. Let equivalent capacitanceofC”and C 4 = C 1 C = 1 C” + 1 C 4 = 1 200 + 1 100 = 2+1 200 = 3 200 C= 200 3 pF Equivalent capacitance of the circuit is 200 3 pF Voltage across C”=V Voltage across C 4 = V 4 V+ V 4 =V =300V Charge oncapacitor C 4 isgivenas: Q 4 =CV= 200 3 × 10 12 ×300 =2× 10 8 C V 4 = Q 4 C 4 = 2× 10 8 100× 10 12 =200V Voltageacrosscapacitor C 1 isgivenbytherelation, V 1 =V V 4 =300200 =100V Voltage V 1 across C 1 is 100 V. Charge on capacitor C 1 , Q 1 = C 1 V 1 =100× 10 12 ×100 = 10 8 C Capacitors C 2 and C 3 having same capacitances have a potential difference of 100 V together. As C 2 and C 3 are in series, the potential difference across C 2 and C 3 is given as, V 2 = V 3 =50 V Charge on C 2 , Q 2 = C 2 V 2 =200× 10 12 ×50 = 10 8 C Charge on C 3 , Q 3 = C 3 V 3 =200× 10 12 ×50 = 10 8 C The equivalent capacitance of the given circuit is 200 3 pFwith Q 2 = C 2 V 2 , Q 3 = C 3 V 3 , V 1 =100V, V 2 = V 3 =50 V, V 4 =200V. 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Q.70 The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.

Ans.

Here,area ofeachplate of a parallel plate capacitor, A=90 cm 2 =90×1 0 4 m 2 Separation between the plates, d=2.5 mm =2.5×1 0 3 m Voltage across the plates, V=400 V ( a ) Capacitance of the capacitor is given as: C= ε 0 A d Electrostatic energy stored in the capacitor is given as: E= 1 2 C V 2 = 1 2 ε 0 A d V 2 Here, ε 0 =Permittivity of free space =8.85 × 1 0 12 C 2 N 1 m 2 E= 1 2 1×8.85×1 0 12 ×90×1 0 4 × ( 400 ) 2 2×2.5×1 0 3 =2.55×1 0 6 J Electrostatic energy stored by the capacitor is 2.55×1 0 6 J. ( b ) Volume of the given capacitorisgivenas: V’=A×d =×90×1 0 4 ×25×1 0 3 =2.25×1 0 4 m 3 Energy stored in the capacitor per unit volume,u= E 1 V’ = 1 2 C V 2 Ad = 1 2 ε 0 A d V 2 Ad = 1 2 ε 0 ( V d ) 2 Here, V d =Electric intensity=E u= 1 2 ε 0 E 2 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Q.71 A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Ans.

Here,Capacitance of the charged capacitor, C 1 =4μF =4× 10 6 F Voltageapplied, V 1 =200 V Electrostatic energy stored in C 1 is given bythe relation, E 1 = 1 2 C 1 V 1 2 = 1 2 ×4× 10 6 × ( 200 ) 2 =8× 10 2 J Capacitance of the uncharged capacitor, C 2 =2μF =2× 10 6 F When C 2 is connected to the circuit, the potential attained =V 2 According to the principleofconservation of charge, Initial charge on capacitor C 1 =Final charge on capacitors, C 1 and C 2 . V 2 ( C 1 + C 2 )= C 1 V 1 V 2 ×( 4+2 )× 10 6 =4× 10 6 ×200 V 2 = 400 3 V Electrostatic energy for the combination of two capacitors is given bytherelation, E 2 = 1 2 ( C 1 + C 2 ) V 2 2 = 1 2 ( 4+2 )× 10 6 × ( 400 3 ) 2 =5.33× 10 2 J Amount of electrostatic energy lost by capacitor C 1 = E 1 E 2 =0.080.0533 =0.0267 J =2.67 × 1 0 2 J 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Q.72

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to 1 2 QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor.

Ans.

Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx. Work done by the force= FΔx As a result, the increaseinpotential energy of the capacitor= uAΔx Here, u = Energy density A = Area of each plate d = Separation between the plates V = Voltage across the plates Work donebyforce= Increase in the potential energy FΔx = uAΔx F = uA = ( 1 2 ε 0 E 2 )A Electric fieldintensity is given as: E = V d E = 1 2 ε 0 ( V d )EA = 1 2 ( ε 0 A V d )E Capacitance,C= ε 0 A d F = 1 2 ( CV )E Charge on the capacitor is given as: Q = CV F = 1 2 QE The physical origin of the factor, 1 2 intheforcecanbe explainedbythefactthat just outsidetheconductor, fieldisEandinsideitiszero. It is the average value, 1 2 of thefield that contributes to the force.

Q.73 A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36). Show that the capacitance of a spherical capacitor is given by

C=4πε0r1r2r1r2where r1and r2are the radii of outer and inner spheres, respectively.

Ans.

Here, radius of outer shell=r1Radius of inner shell=r2Charge onthe inner surface of the outer shell=+Q.Charge induced on the outer surface of the inner shell=Q.Potential difference between the two shells is given as:V=Q4πε0r2Q4πε0r1Here,ε0= Permittivity of free spaceV=Q4πε0[1r21r1]=Q(r1r2)4πε0r1r2Capacitanceofthegivensystemisgivenas:C=Charge(Q)Potential difference(V)=4πε0r1r2r1r2Capacitanceofthegivensphericalcapacitoris4πε0r1r2r1r2.

Q.74 A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 μC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere?
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.

Ans.

Here,radius of inner sphere, r a =12 cm=0.12 m Radius ofouter sphere, r b =13 cm=0.13 m Charge on inner sphere,q=2.5μC=2.5× 10 6 C Dielectric constant of theliquid, ε r =32 (a)Capacitanceofthecapacitorisgivenbythe relation, C= 4π ε 0 ε r r a r b r b r a Here, ε 0 = Permittivity of free space =8.85× 10 12 C 2 N 1 m 2 1 4π ε 0 =9× 10 9 N m 2 C 2 C= 32×0.12×0.13 9× 10 9 ×( 0.130.12 ) 5.5× 10 9 F The capacitance of the capacitor is approximately 5.5× 10 9 F. ( b ) Potential of the inner sphere is given bythe relation, V= q C = 2.5× 10 6 5.5× 10 9 =4.5× 10 2 V The potential of the inner sphere is 4.5× 10 2 V. ( c ) Here,radius of an isolated sphere, r=12×1 0 2 m Capacitance of anisolatedsphere,C’=4π ε 0 r =4π×8.85× 10 12 ×12×1 0 2 =1.33× 10 11 F Capacitance of an isolated sphere is muchless in comparison to the concentric spheres. ∵Intheconcentric spheresthe outer sphere is earthed. Potential difference isless and the capacitance is more than thatofanisolated sphere. 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Q.75 Answer carefully:
(a) Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q1Q2/4πr 2, where r is the distance between their centres?
(b) If Coulomb’s law involved 1/r3 dependence (instead of 1/r2), would Gauss’s law be still true?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?
(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?
(f) What meaning would you give to the capacitance of a single conductor
(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

Ans.

(a) When two charged spheres are brought close to each other, the charge distribution on them does not remain uniform. Therefore, the force between two conducting spheres is not exactly given by the given expression.
(b) Gauss’s law will not be true, if Coulomb’s law involved 1/r3 dependence, instead of 1/r2 dependence.
(c) Yes, if a small test charge is released at rest at a point in an electrostatic field configuration, then it will move along the line of force passing through that point, only if the field lines are straight. If the field lines are not straight, the charge will not go along the line. This is because the field lines give the direction of acceleration.
(d)The direction of force due to field is towards the nucleus, and the electron does not move along the direction of this force. Therefore, whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.
(e) No, electric potential is continuous across the surface of a charged conductor.
(f) The capacitance of a single conductor implies a parallel plate capacitor with one of its two plates at infinity.
(g) A water molecule has an unsymmetrical shape as compared to that of mica. Therefore, it has a permanent dipole moment. That is why; it has a greater dielectric constant than mica.

Q.76 A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 5C. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).

Ans.

Here,length of coaxial cylinders, l=15 cm =0.15 m Radius of theouter cylinder, r 1 =1.5 cm =0.015 m Radius of theinner cylinder, r 2 =1.4 cm =0.014 m Charge on the inner cylinder, q=3.5 μC = 3.5×1 0 6 C Capacitanceofthesystemisgivenbytherelation, C= 2π ε 0 l lo g e r 1 r 2 Here, ε 0 = Permittivity of free space =8.85× 10 -12 N 1 m 2 C 2 C= 2π×8.85× 10 -12 ×0.15 2.3026 log 10 ( 0.15 0.14 ) = 2π×8.85× 10 -12 ×0.15 2.3026×0.0299 =1.2× 10 -10 F Potential difference of the inner cylinder,V= q C = 3.5× 10 6 1.2× 10 -10 =2.92× 10 4 V MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@EBD0@

Q.77 A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?

Ans.

Here,voltage rating of a parallel plate capacitor, V =1 kV=1000 V Dielectric constantofthematerial, ε r =3 Dielectric strength=1 0 7 Vm -1 For safety, the electricfield intensity shouldnever exceed10% of the dielectric strength. Electricfield,E=10% of1 0 7 = 10 6 V m 1 Capacitance of parallel plate capacitor, C=50 pF =50×1 0 12 F Distance between the plates is givenbytherelation, d= V E = 1000 10 6 = 10 3 m Capacitanceisgivenas: C= ε 0 ε r A d Here,A=Area of each plateofthecapacitor ε 0 = Permittivity of free space =8.85×1 0 12 N 1 C 2 m 2 A= C ε 0 ε r = 50×1 0 12 8.85×1 0 12 ×3 19c m 2 The area of each plate is approximately 19 cm 2 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@63B3@

Q.78 Describe schematically the equipotential surfaces corresponding to
(a) a constant electric field in the z-direction,
(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.

Ans.

(a) The equipotential surfaces are the equidistant planes parallel to the x-y plane.
(b) The equipotential surfaces are the planes parallel to the x-y plane. When the field increases uniformly, planes get closer (differing by fixed potential).
(c) The equipotential surfaces are the concentric spheres centered at the origin.
(d) The equipotential surfaces have the periodically varying shape. The shape of equipotential surfaces becomes parallel to the grid at a far off distance.

Q.79 In a Van de Graaff type generator a spherical metal shell is to be a 15 × 106 V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 107 Vm−1. What is the minimum radius of the spherical shell required?

(You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)

Ans.

Here,voltage, V=15×1 0 6 V Dielectric strength=5×1 0 7 Vm -1 Electric field, E=Dielectric strength =5×1 0 7 Vm -1 Since,E= V r Minimum radius of the spherical shell,r= V E = 15×1 0 6 5×1 0 7 =3cm The minimum radius of the spherical shell required is 30 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@0B0E@

Q.80 A small sphere of radius r1 and charge q1 is enclosed by a spherical shell of radius r2 and charge q2. Show that if q1 is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2 on the shell is.

Ans.

Thepotential of outer sheprical shell, V2 =[ Potential due to its own charge q2 + potential due to inner charge q1 ] Potential of the inner solid shpere, V1 =[ Potential due to its own charge q1 + potential due to inner charge q2 ] V1V2= q1 4πεo [ 1 r1 1 r2 ]; which is a positive quantity MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@2DFA@

As the charge lies always on the outer surface of the shell, therefore, when the sphere and the shell are connected by a wire, charge will flow from the sphere to the shell, whatever be the magnitude and sign of charge q2.

Q.81 Answer the following:
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm−1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)
(b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m2. Will he get an electric shock if he touches the metal sheet next morning?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about 100 Vm−1 at its surface in the downward direction, corresponding to a surface charge density = −10−9 Cm−2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)

Ans.

(a) As we step out of our house, we do not get an electric shock. This is because the original equipotential surfaces of open air change keeps our body and the ground at the same potential.
(b) Yes, the man will get an electric shock because the steady discharging current in the atmosphere charges up the aluminium sheet and raises its voltage gradually. The rise in its voltage depends on the capacitance of the capacitor formed by the aluminium sheet and the ground.
(c) The atmosphere is charged continuously by occurrence of thunderstorms and lightning. Therefore, even with the presence of discharging current of 1800 A in the atmosphere, the atmosphere is not discharged completely. The two opposing currents are, on an average, in equilibrium and the atmosphere remains electrically charged.
(d) During a lightning, the electrical energy of the atmosphere is dissipated in the form of light energy, heat energy, and sound energy.

Q.82 The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

Ans.

Here,emf of the battery, E = 12 V Internal resistance of the battery, r = 0.4 Ω Letmaximum current drawnfromthebattery = I According to Ohm’s law, E = Ir I = E r = 12 V 0.4 Ω = 30 A Maximum current drawn from the given battery is 30 A.

Q.83 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Ans.

Here,emf of battery, E = 10 V Internal resistance ofbattery, r = 3 Ω Currentflowing in the circuit, I = 0.5 A Letresistance of the resistor = R From Ohm’s law,the relation for current isgivenby, I = E R+r R+r = E I = 10V 0.5A = 20Ω R = 20Ω – 3Ω = 17Ω Letterminal voltage of the resistor = V From Ohm’s law,wehave V = IR V = 0.5 A × 17 Ω = 8.5 V The resistance of the resistor is 17 Ωand the terminal voltage is8.5 V.

Q.84 (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Ans.

(a) Here, R 1 = 1 Ω R 2 = 2 Ω R 3 = 3 Ω Asthe resistorsare combined in series,total resistance of the combination is equalto the algebraic sum of individual resistances. Total resistance, R s = R 1 + R 2 + R 3 = 1 Ω + 2 Ω + 3 Ω = 6Ω (b) Here,emf of the battery, E = 12 V Letcurrent flowing through the circuit = I Total resistance of circuit, R = 6 Ω Using Ohm’s law,therelationforcurrentisgivenas, I = E R = 12 V 6 Ω = 2A UsingOhm’slaw, Potential drop across resistor R 1 isgivenby, V 1 = IR 1 = 2 A×1 Ω = 2V Potential drop across resistor R 2 isgivenby, V 2 = IR 2 = 2 A×2 Ω = 4V Potential drop across resistor R 3 isgivenby, V 3 = IR 3 = 2 A×3 Ω = 6V The potential drop across resistors R 1 , R 2 and R 3 are 2 V, 4 V, and 6 Vrespectively.

Q.85 (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

Ans.

(a) Here, R 1 = 2 Ω R 2 = 4 Ω R 3 = 5 Ω Astheresistors are connected in parallel, Total resistance of the combination is givenbytherelation, 1 R = 1 R 1 + 1 R 2 + 1 R 3 1 R = 1 2 Ω + 1 4 Ω + 1 5 Ω = 10+5+4 20 Ω = 19 20 Ω R= 20 19 Ω Total resistance of the combination is 20 19 Ω. (b) Here,emf of the battery, V = 20 V UsingOhm’slaw Current through resistor R 1 is given by, I 1 = V R 1 = 20 V 2 Ω = 10A Current through resistor R 2 is given by, I 2 = V R 2 = 20 V 4 Ω = 5A Current through resistor R 3 is given by, I 3 = V R 3 = 20 V 5 Ω = 4A Total current, I = I 1 +I 2 +I 3 = 10 A + 5 A + 4 A = 19 A The current flowingthrough resistors R 1 , R 2 and R 3 is10 A,5 A, and 4 A respectively and thetotal current is 19 A.

Q.86 At room temperature (27.0°C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10-4°C-1.

Ans.

Here,room temperature, T = 27°C Resistance ofheating element at T, R = 100 Ω Let increased temperature of the element = T 1 Resistance of heating element at T 1 , R 1 = 117 Ω Temperature co-efficient of the material of the resistor,α = 1 .70×10 -4 ° C -1 Therelationforαisgivenas, α = R 1 – R R( T 1 – T ) T 1 – T = R 1 – R Rα T 1 – 27°C = 117 Ω -100 Ω 100 Ω×( 1 .70×10 -4 ° C -1 ) T 1 – 27°C = 1000°C T 1 = 1000+27°C = 1027°C At 1027°C, the resistance of the element is 117 Ω.

Q.87 A negligibly small current is passed through a wire of length 15 m and uniform cross section 6.0 × 10−7 m2, and its resistance is measured to be 5.0 Ω.What is the resistivity of the material at the temperature of the experiment?

Ans.

Here,length of wire, l = 15 m Cross-sectionalarea of the wire, a = 6 .0×10 -7 m 2 Resistance ofthe material of wire, R = 5.0 Ω Letresistivity ofthematerial of wire = ρ Resistance is relatedto resistivity bytherelation, R=ρ l A ρ= RA l ρ= 5.0 Ω×6 .0×10 -7 m 2 15 m = 2×10 -7 Ωm The resistivity of the material is 2 × 10 -7 Ωm.

Q.88 A silver wire has a resistance of 2.1 Ω at 27.5°C, and a resistance of 2.7 Ω at 100°C. Determine the temperature coefficient of resistivity of silver.

Ans.

Here, Temperature, T 1 = 27.5°C Resistance of silver wire at T 1 , R 1 = 2.1 Ω Temperature, T 2 = 100°C Resistance of silver wire at T 2 , R 2 = 2.7 Ω Lettemperature coefficient of silver be α αisrelatedwith temperature and resistance as: α = R 2 – R 1 R 1 ( T 2 – T 1 ) = 2.7 Ω – 2.1 Ω 2.1 Ω×( 100°C27.5°C ) = 0 .0039°C -1 The temperature coefficient of silver is 0.0039 °C -1 .

Q.89 A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0°C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10−4°C−1.

Ans.

Here,voltageapplied, V=230 V Initial current, I 1 =3.2 A Let initial resistance be R 1 UsingOhm’slaw, R 1 canbecalculatedas: R 1 = V I 1 R 1 = 230V 3.2 A = 71.87Ω Steady state value of current, I 2 = 2.8 A Resistance at the steady state isgivenby, R 2 = V I 2 R 2 = 230 V 2.8 A = 82.14Ω Temperature co-efficient of resistanceofnichrome, α=1 .70×10 -4 ° C -1 Initial temperature of nichrome, T 1 = 27.0°C Let steady state temperature of nichrome = T 2 T 2 can be calculated by usingthe relation for α, α= R 2 – R 1 R 1 ( T 2 – T 1 ) T 2 – T 1 = R 2 – R 1 α R 1 T 2 – 27.0°C = 82.14 Ω– 71.87 Ω 1 .70×10 -4 ° C -1 ×71.87 Ω = 840.5°C T 2 = 840.5°C + 27°C = 867.5°C The steady temperature of the heating element is 867.5°C.

Q.90 Determine the current in each branch of the network shown in fig 3.30:

Ans.

The current flowing through various branches of the circuit is shown in the given figure.

Here, I1 = Current passing through outer circuit

I2 = Current passing through arm AB

I3 = Current passing through arm AD

I2 − I4 = Current passing through branch BC

I3 + I4 = Current passing through branch CD

I 4 = Current passing through branch BD For closed circuit ABDA, potential is zero i.e., 10 I 2 + 5I 4 5I 3 = 0 2I 2 + I 4 I 3 = 0 I 3 = 2I 2 + I 4 ( 1 ) For closed circuit BCDB, potential is zero i.e., 5( I 2 I 4 ) 10( I 3 + I 4 ) 5I 4 = 0 5I 2 + 5I 4 10 I 3 10 I 4 5I 4 = 0 5I 2 10 I 3 20 I 4 = 0 I 2 = 2I 3 + 4I 4 ( 2 ) For closed circuit ABCFEA, potential is zero i.e., 10 + 10 ( I 1 ) + 10( I 2 ) + 5( I 2 I 4 ) = 0 10 = 15I 2 + 10 I 1 5I 4 3I 2 + 2I 1 I 4 = 2 ( 3 ) From equations ( 1 ) and ( 2 ), we get I 3 = 2( 2I 3 + 4I 4 ) + I 4 I 3 = 4I 3 + 8I 4 + I 4 3I 3 = 9I 4 3I 4 = + I 3 ( 4 ) Substituting equation ( 4 ) in equation ( 1 ), we have I 3 = 2I 2 + I 4 4I 4 = 2I 2 I 2 = 2I 4 ( 5 ) From the given figure,it is clear that, I 1 = I 3 + I 2 ( 6 ) Substituting equation ( 6 ) in equation ( 1 ), we have 3I 2 +2( I 3 + I 2 ) I 4 =2 5I 2 + 2I 3 I 4 =2 ( 7 ) Substituting equations ( 4 ) and ( 5 ) in equation ( 7 ), we have 5( 2I 4 )+2( 3I 4 ) I 4 =2 10 I 4 6I 4 I 4 =2 17 I 4 =2 I 4 = 2 17 A Equation ( 4 ) becomes I 3 = 3( I 4 ) = 3( 2 17 )= 6 17 A I 2 = 2( I 4 ) = 2( 2 17 ) = 4 17 A I 2 I 4 = 4 17 ( 2 17 ) = 6 17 A I 3 + I 4 = 6 17 +( 2 17 ) = 4 17 A I 1 = I 3 + I 2 = 6 17 + 4 17 = 10 17 A Currentin branch CD = -4 17 A Currentin branch AD = 6 17 A Currentin branch BD = ( -2 17 )A Total current =( 4 17 + 6 17 + -4 17 + 6 17 + -2 17 )A = 10 17 A

Q.91 (a) In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?
(b) Determine the balance point of the bridge above if X and Y are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

Ans.

The given figure shows a metre bridge with resistors X and Y.

(a) Here, balance point from end A, l1 = 39.5 cm

Resistance of resistor Y = 12.5 Ω

Thecondition for the balance is given as, X Y = 100 – l 1 l 1 The resistance of resistor X is 8.2 Ω. The connections between resistors in a Wheatstone or metre bridge are made of thick copper strips to minimize the resistanceofconnections, which is not accounted in the bridge formula.

b) As X and Y are interchanged, therefore l1 and (100 − l1) also get interchanged.

Balance point of the meter bridge will be (100 − l1) from A.

100 cm − l1 = 100 cm − 39.5 cm = 60.5 cm

Hence, balance point of the meter bridge is 60.5 cm from A.

(c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will not show deflection. Hence, the galvanometer will show no current.

Q.92 A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Ans.

Here, emf of the battery, E = 8.0 V

Internal resistance of battery, r = 0.5 Ω

Voltage of DC supply, V = 120 V

Resistance of external resistor, R = 15.5 Ω

Let effective voltage in the circuit = V1

As R is connected to the storage battery in series,

V 1 = V E V 1 = 120 V 8 V = 112 V Current flowing in the circuit(I) given by the relation, I = V 1 R+r I = 112 V 15.5 Ω + 5 Ω = 7A Voltage across resistor R given as: IR = 7 A×15.5 Ω = 108.5 V DC supply voltage = Terminal voltage of battery + Voltage drop acrossresistor R Terminal voltage of battery =120 V – 108.5 V = 11.5 V

A series resistor in a charging circuit limits the current drawn from the external source of d.c. supply. In its absence, the current will be extremely high. This can be very dangerous.

Q.93 In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

Ans.

Here, emf of cell, E 1 = 1.25 V Balance point of potentiometer, l 1 = 35 cm As the cell is replaced by another cell of emf E 2 , New balance point of potentiometer, l 2 = 63 cm Theconditionforbalanceisgivenbytherelation, E 1 E 2 = l 1 l 2 E 2 = E 1 × l 2 l 1 E 2 = 1.25 V× 63 cm 35 cm = 2.25V Emf of the second cell is 2.25V.

Q.94 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 m−3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross- section of the wire is 2.0 × 10−6 m2 and it is carrying a current of 3.0 A.

Ans.

Here,number density of free electrons in copper conductor, n = 8 .5×10 28 m -3 Length of copper wire, l = 3.0 m Cross-sectionalarea of the wire, A = 2 .0×10 -6 m 2 Current flowingthrough the wire, I = 3.0 A, Current flowingthrough the wireis given by the relation, I = nAeV d Here,Electric charge,e = 1 .6×10 -19 C V d = Drift velocity I = nAeV d Since, V d = lengthofwire(l) timetakentocoverl(t) I = nAe l t t = nAel I = 3 m×8 .5×10 28 m -3 ×2 .0×10 -6 m 2 ×1 .6×10 -19 C 3 A = 2 .7×10 4 s The time taken by an electron to drift from one end of the wire to the other is2 .7 × 10 4 s.

Q.95 The earth’s surface has a negative surface charge density of 10−9 Cm−2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 106 m.)

Ans.

Here,surface charge density of the earth’s surface, σ = 10 -9 Cm -2<