# Class 12 NCERT Solutions for Physics Chapter 1

## Class 12 Physics Chapter 1 NCERT Solutions – Electric Charge & Field

NCERT Class 12 Physics Chapter 1 is all about electric charges and electric fields. Physics is a vast subject with a lot of information yet to be explored and hence needs expertise and experience on the part of the teacher to impart that knowledge to the students. While classroom teaching covers a part of this knowledge transfer, the rest falls upon the students. To help students soak in this knowledge, Extramarks experts compiled the class 12 physics chapter 1 NCERT Solutions.

All the content related to Class 12 Physics Chapter 1 NCERT Solutions is available on Extramarks. The subject matter experts at Extramarks have created the solution sets based on their knowledge expertise and NCERT guidelines. Furthermore, all the questions listed in the NCERT Physics textbook have been answered here. In addition, all the solutions, syllabus, exemplars and important questions are available on Extramarks.

### Key Topics Covered In NCERT Solutions Class 12 Physics Chapter 1

Class 12 Physics Chapter 1 NCERT Solutions is based on Electric Charge and Field. Following are the key topics in this chapter.

 Ex 1.1 Introduction Ex 1.2 Electric Charge Ex 1.3 Conductors And Insulators Ex 1.4 Charging By Induction Ex 1.5 Basic Properties Of Electric Charge Ex 1.5.1 Additivity Of Charges Ex 1.5.2 Charge Is Conserved Ex 1.5.3 Quantisation Of Charge Ex 1.6 Coulomb’S Law Ex 1.7 Forces Between Multiple Charges Ex 1.8 Electric Field Ex 1.8.1 Electric Field Due To A System Of Charges Ex 1.8.2 Physical Significance Of Electric Field Ex 1.9 Electric Field Lines Ex 1.10 Electric Flux Ex 1.11 Electric Dipole Ex 1.11.1 The Field Of An Electric Dipole Ex 1.11.2 Physical Significance Of Dipoles Ex 1.12 Dipole In A Uniform External Field Ex 1.13 Continuous Charge Distribution Ex 1.14 Gauss’S Law Ex 1.15 Applications Of Gauss’S Law Ex 1.15.1 Electric Field Due To An Infinitely Long Straight Uniformly Charged Wire Ex 1.15.2 Electric  Field Due To A Uniformly Charged Infinite Plane Sheet Ex 1.15.3 Electric Field Due To A Uniformly Charged Thin Spherical Shell

NCERT Solutions for Class 12 Physics Chapter 1 covers many concepts revolving around Electric charge and electric field. It also introduces electric field topics such as Flux, Dipole, and various laws like Gauss’Law, Coulomb’s Law.

The weightage of the CBSE NCERT Solutions for Class 12 Physics Chapter 1 can help students score a maximum of 15 marks. If students face difficulties understanding the concepts or cannot figure out the solutions, they may log on to Extramarks to get expert guidance within minutes.

Here’s detailed information on each subtopic in NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges

### 1.2 Electric Charge

While exercise 1.1 gives a brief introduction to the chapter, exercise 1.2 of the NCERT Solutions for Class 12 Physics Chapter 1  gives more details on electric charge. The word “electricity” means “amber.” The magnetic and electric forces in materials, atoms, and molecules affect their characteristics. The word “electric charge” refers to the two types of entities.

According to an experiment, there are two types of electrification: (i) like charges repel one another, and (ii) different charges attract each other. The polarity of charge distinguishes between the two types of charges.

### 1.3 Conductors and Insulators:

During the conduction experiment, it was stated that conductors help in the flow of electric charge, but insulators do not. All types of metals, Earth, and Human Bodies are all conductors, but porcelain, nylon, and wood are all insulators, offering high resistance to the flow of electricity through them.

### 1.5 Basic Properties Of Electric Charge

There are three essential qualities of an electric charge mentioned under NCERT Solutions for Class 12 Physics Chapter 1.

• Charge – According to these characteristics, the total charge of a body indicates the integral multiple of a fundamental quantum of charge.
• Additive- This electric charge attribute reflects a body’s overall charge as the algebraic sum of all singular charges acting on the system.
• Conservation- According to conservation, the entire system’s charge remains unchanged. In other words, when things get charged due to friction, the charge is transferred from one item to another. Charges can neither be created nor destroyed.

### 1.6 Coulomb’s Law

Under this section of NCERT Solutions for Class 12 Physics Chapter 1, students learn about Coulomb’s Law. Coulomb’s law states that the mutual electrostatic force existing between two point charges A and B is proportional to their product AB and inversely proportional to the square of the distance between them.

The equation is FAB = force on B due to A = k(AB) r2AB.

The value of k is 9109Nm2C-2 and is termed a constant of proportionality.

### 1.7 Multiple Charges and Their Forces

Coulomb’s law computes the mutual electric force between two charges. However, it cannot determine the force on a charge when several charges are present. An experiment has shown that the vector sum of all the forces on any charge owing to many other charges, taken one at a time, equals the vector sum of all the forces on that charge due to the other charges.

This section also talks about the Principles of Superposition: The property of two charges to repel and attract each other is unaffected by a third extra charge, according to the superposition principle. Students may refer to NCERT Solutions Class 12 Physics Chapter 1.7 to know more.

### 1.10 Electric Flux

The wide variety of electrical area traces passing a given vicinity in a unit of time is described as the electrical flux. However, we notice that there’s no float of an observable bodily amount like withinside the case of liquid float. As per definition, Electric flux Δθ via a place detail ΔS is described by

Δθ= E.ΔS= E ΔS cosθ

This is proportional to the variety of area traces reducing the vicinity detail. The altitude θ is the altitude between E and ∆S. On a closed floor, where the conference is stated, θ is the altitude among E and the outward regular to the vicinity detail. To calculate the total flux via any given floor, divide the floor into small vicinity elements, calculate the flux at every point and sum them. Thus, the full flux θ via a floor S is θ ~ Σ E. ∆S. The usage of approximation image is because the electrical area E is to be steady over the small vicinity detail.

### 1.11 Electric Dipole

A couple of same or contrary fees A and -B are separated using distance 2x. The dipole second vector (let’s expect it as p) has an importance 2Ax and is within the dipole axis’s route from -B to A. To know more about Electric Dipole, students may refer to NCERT Solutions Class 12 Physics Chapter 1.11.

### 1.14 What is Gauss’Law?

According to the Gauss regulation, the overall flux related to a closed floor is 1/ε0 instances of the price enclosed using the closed floor.

E.ds=10q

For example, a factor price q is located inside a die of edge’ a’. As in keeping with Gauss regulation, the flux through every face of the die is q/6ε0.

The electric-powered subject is the simple idea to recognise approximately electricity. Generally, the electrical subject of the floor is calculated via means of making use of Coulomb’s regulation. However, to estimate the electrical subject distribution on a closed floor, we want to apprehend the idea of Gauss regulation. It explains the electrical price enclosed in a closed or the electrical price gift withinside the enclosed closed floor.

### Gauss’ Law Formula

In keeping with the Gauss theorem, the overall price enclosed in a closed floor is proportional to the total flux enclosed using the floor. Therefore, if ϕ is overall flux and ϵ0 is electrical constant, the general electric powered price Q surrounded by way of means of the floor is:

Q = ϕ ϵ0.

The Gauss regulation components are expressed by means of ϕ = Q/ϵ0.

Where,

Q = overall price on the given floor,

ε0 = the electrical constant.

For more information on the Gauss Law, students may access NCERT Solutions Class 12 Physics Chapter 1 on Extramarks.

The Gauss’ Theorem

The net flux through a closed floor is proportional to the net price within the quantity enclosed using the closed floor.

Φ = → E.d → A = qnet/ε0

In easy words, the Gauss theorem relates the ‘flow’ of electrical subject strains (flux) to the expenses on the enclosed floor. If there aren’t any expenses enclosed using a floor, then the net electric-powered flux stays zero. This approach shows that the quantity of electrical subject strains to come into the floor is the same as the sphere strains leaving the floor.

### 1.15 Applications of Gauss Law

Gauss Law is applied.

• For a charged ring with Radius R, which is located on-axis, at a distance x from the centre of the ring:

E=140.qx(R2+x2)3/2

At the centre, x = zero and E = zero.

• In the case of a limitless price line, at a distance, ‘r’. E = (1/4 × πrε0) (2π/r) = λ/2πrε0. Where λ is the linear price density.
• The depth of the electrical subject close to an aircraft sheet of price is E = σ/2ε0K, wherein σ = floor price density.
• The depth of the electrical subject is close to an aircraft charged conductor E = σ/Kε0 in a medium of dielectric steady K. If the dielectric medium is air, then EAir = σ/ε0.
• The subject among parallel plates of a condenser is E = σ/ε0, wherein σ is the floor price density.

Students may click on the respective topics of the NCERT Solutions Class 12 Physics Chapter 1 in the table to access the study material provided by Extramarks.

### NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields NCERT Solutions Article Links

Click on the below links to view NCERT Solutions Class 12 Physics Chapter 1:

• Class 12 Physics Chapter 1: Exercise 1.1
• Class 12 Physics Chapter 1: Exercise 1.2
• Class 12 Physics Chapter 1: Exercise 1.3
• Class 12 Physics Chapter 1: Exercise 1.4
• Class 12 Physics Chapter 1: Exercise 1.5
• Class 12 Physics Chapter 1: Exercise 1.6
• Class 12 Physics Chapter 1: Exercise 1.7
• Class 12 Physics Chapter 1: Exercise 1.8
• Class 12 Physics Chapter 1: Exercise 1.9
• Class 12 Physics Chapter 1: Exercise 1.10
• Class 12 Physics Chapter 1: Exercise 1.11
• Class 12 Physics Chapter 1: Exercise 1.12
• Class 12 Physics Chapter 1: Exercise 1.13
• Class 12 Physics Chapter 1: Exercise 1.14
• Class 12 Physics Chapter 1: Exercise 1.15

Students may access NCERT Solutions Class 12 Physics Chapter 1 as well as other chapters by clicking here. In addition, students can also explore NCERT Solutions for other classes below.

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By referring to Extramarks NCERT Solutions Class 12 Physics Chapter 1, students can easily understand Electric Charge and Field. In addition to Solutions, students may refer to past years’ question papers, important questions and revision notes on Extramarks.

### Key Features of Class 12 Physics Ch 1 NCERT Solutions

Physics is an important subject for any competitive examinations like JEE or NEET. The subject cannot be memorised and needs a detailed approach to every problem. And hence students must study this subject by understanding the concepts well. NCERT Solutions Class 12 Physics Chapter 1 gives you in-depth knowledge about the subject and concepts to be studied.

Students searching for Class 12 Physics Ch 1 NCERT Solutions can access it on Extramarks. Some of the major reasons you must choose Extramarks include

• The solutions are prepared by experts in the field of physics.
• The Solution is as per the updated NCERT guidelines.
• All solutions are mentioned in a detailed and easy-to-understand manner.
• Every detail provided is framed considering the student’s learning capacity.
• Students can score well through these explanatory NCERT Solutions Class 12 Physics Chapter 1 provided by Extramarks.

Q.1 What is the force between two small charged spheres having charges of 2 × 10−7 C and 3 × 10−7 C placed 30 cm apart in air?

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}c\text{harge on first sphere},{\text{q}}_{\text{1}}=\text{2}×\text{1}{0}^{-\text{7}}\text{C}\\ \text{Charge on second sphere},{\text{q}}_{\text{2}}=\text{3}×\text{1}{0}^{-\text{7}}\text{C}\\ \text{Distance between spheres},\text{r}=\text{3}0\text{cm}\\ =0.\text{3 m}\\ \text{Electrostatic force between the spheres is given as:}\\ \text{F=}\frac{{q}_{1}{q}_{2}}{4\pi {\epsilon }_{0}{r}^{2}}\\ \text{Here},{\in }_{0}=\text{Permittivity of free space}\\ \frac{1}{4\pi {\epsilon }_{0}}=9×{10}^{9}\text{\hspace{0.17em}}N{m}^{2}{C}^{-2}\\ \therefore F=\frac{9×{10}^{9}×\text{2}×\text{1}{0}^{-\text{7}}×\text{3}×\text{1}{0}^{-\text{7}}}{{\left(0.3\right)}^{2}}\\ =6×{10}^{-3}\text{\hspace{0.17em}}N\\ \therefore \text{Force between the two small charged spheres}\\ \text{=6}×\text{1}{0}^{-\text{3}}\text{N}.\text{}\\ \text{Since}\text{\hspace{0.17em}}t\text{he charges are}\text{\hspace{0.17em}}\text{of same nature,}\\ \therefore \text{Force between them will be repulsive}.\end{array}$

Q.2 The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{Here,}\text{\hspace{0.17em}}\text{Electrostatic force on first sphere},\text{\hspace{0.17em}}\text{F}=\text{}0.\text{2 N}\\ \text{Charge on first sphere},{\text{q}}_{\text{1}}=0.\text{4}\mu \text{C}\\ =0.\text{4}×\text{1}{0}^{-\text{6}}\text{C}\\ \text{Charge on second sphere},{\text{q}}_{\text{2}}=-0.\text{8}\mu \text{C}\\ =-0.\text{8}×\text{1}{0}^{-\text{6}}\text{C}\\ \text{Electrostatic force between the spheres is given as:}\\ \text{F=}\frac{{q}_{1}{q}_{2}}{4\pi {\epsilon }_{0}{r}^{2}}\\ \text{Here},\text{\hspace{0.17em}}{\epsilon }_{0}=\text{Permittivity of free space}\\ \frac{1}{4\pi {\epsilon }_{0}}=9×{10}^{9}\text{\hspace{0.17em}}N{m}^{2}{C}^{-2}\\ \therefore {r}^{2}=\frac{{q}_{1}{q}_{2}}{4\pi {\epsilon }_{0}F}\\ =\frac{9×{10}^{9}×0.4×\text{1}{0}^{-6}×0.8×\text{1}{0}^{-6}}{0.2}\\ =144×{10}^{-4}\text{\hspace{0.17em}}N\\ r=\sqrt{144×{10}^{-4}}=0.12\text{\hspace{0.17em}}m\\ \therefore \text{Distance between the two spheres =}0.12\text{m}.\\ \left(\text{b}\right)\text{As}\text{\hspace{0.17em}}b\text{oth the spheres attract each other with}\\ \text{same force,}\\ \therefore F\text{orce on}\text{\hspace{0.17em}}\text{the second sphere due to the first sphere}\text{\hspace{0.17em}}\\ \text{is}0.\text{2 N}.\end{array}$

Q.3 Check that the ratio ke2/Gmemp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Ans.

$\begin{array}{l}\text{The given ratio =}\frac{k{e}^{2}}{G{m}_{e}{m}_{p}}\\ \text{Here},\text{G}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{the Gravitational constant,}\text{\hspace{0.17em}}its\text{\hspace{0.17em}}{\text{unit=Nm}}^{\text{2}}{\text{kg}}^{-\text{2}}\\ {\text{m}}_{\text{e}}{\text{and m}}_{\text{p}}\text{are}\text{\hspace{0.17em}}\text{masses of electron and proton,}\text{\hspace{0.17em}}\text{their}\text{\hspace{0.17em}}\\ \text{unit= kg}.\\ \text{e is}\text{\hspace{0.17em}}\text{the electric charge,}\text{\hspace{0.17em}}i\text{ts unit= C}\\ \text{K=}\text{\hspace{0.17em}}\text{A}\text{\hspace{0.17em}}\text{constant=}\frac{1}{4\pi {\epsilon }_{0}}\\ {\epsilon }_{0}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}the\text{\hspace{0.17em}}p\text{ermittivity of free space,}\text{\hspace{0.17em}}i\text{ts unit=}\text{\hspace{0.17em}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\\ \therefore The\text{\hspace{0.17em}}given\text{\hspace{0.17em}}ratio,\text{\hspace{0.17em}}\frac{k{e}^{2}}{G{m}_{e}{m}_{p}}=\frac{\left[{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\right]\left[{\text{C}}^{-\text{2}}\right]}{\left[{\text{Nm}}^{\text{2}}{\text{kg}}^{-\text{2}}\right]\left[\text{kg}\right]\left[\text{kg}\right]}\\ ={M}^{0}{L}^{0}{T}^{0}\\ \therefore \text{The given ratio is dimensionless}.\\ \text{e = 1}{\text{.6 × 10}}^{\text{-19}}\text{C}\\ \text{G = 6}{\text{.67 × 10}}^{\text{-11}}{\text{N m}}^{\text{2}}{\text{kg}}^{\text{-2}}\\ {\text{m}}_{\text{e}}\text{= 9}{\text{.1 × 10}}^{\text{-31}}\text{kg}\\ {\text{m}}_{\text{p}}\text{= 1}{\text{.66 × 10}}^{\text{-27}}\text{kg}\\ \therefore N\text{umerical value of the given ratio}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\text{\hspace{0.17em}}\\ \frac{k{e}^{2}}{G{m}_{e}{m}_{p}}=\frac{9×{10}^{9}×{\left(\text{1}.\text{6}×\text{1}{0}^{-\text{19}}\right)}^{2}}{\text{6}.\text{67}×\text{1}{0}^{-\text{11}}×\text{9}.\text{1}×\text{1}{0}^{-\text{31}}×\text{1}.\text{66}×\text{1}{0}^{-\text{27}}}\\ \approx 2.3×{10}^{39}\\ \text{This is the ratio of electrostatic force to gravitational}\\ \text{force between an}\text{\hspace{0.17em}}\text{electron and a proton,}\text{\hspace{0.17em}}\text{keeping}\\ \text{distance between them constant}.\end{array}$

Q.4 (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Ans.

(a) Electric charge of a body is quantized means that only integral (1, 2, …., n) number of elementary charge can be transferred from one body to the other. The elementary charge is charge on an electron or a proton which cannot be transferred in fraction.
(b) At macroscopic level, the charges used are enormous as compared to the magnitude of electric charge. Thus, quantization of charge can be ignored and can be considered as continuous charge.

Q.5 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.6 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. $\begin{array}{l}\text{In}\text{\hspace{0.17em}}\text{the given figure,}\text{\hspace{0.17em}}\text{there}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{a}\text{\hspace{0.17em}}\text{square of side 1}0\text{cm with}\\ \text{four charges placed at its corners}.\text{}\\ \text{S is the centre of the square}.\\ \text{where},\\ \text{Length}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{side}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{square=AB}=\text{BC}=\text{CD}=\text{AD}=\text{1}0\text{cm}\\ \text{Length}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{diagonal}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{square=AC}=\text{BD}=\text{10}\sqrt{2}\text{\hspace{0.17em}}\text{cm}\\ \text{AS}=\text{SC}=\text{DS}=\text{SB}=5\sqrt{2}\text{cm}\\ \text{A charge of amount 1}\mu \text{C is placed at the}\text{\hspace{0.17em}}centre\text{\hspace{0.17em}}S\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\\ the\text{\hspace{0.17em}}square.\end{array}$

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.7 (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.8 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. $\begin{array}{l}\left(\text{a}\right)\text{The situation}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{problem is shown in the given}\\ \text{figure}.\text{}\\ \text{Here,}\text{\hspace{0.17em}}\text{O is the mid}-\text{point of line AB}.\\ {\text{q}}_{\text{A}}\text{=3μC}\\ {\text{=3×10}}^{\text{-6}}\text{\hspace{0.17em}}\text{C}\\ {\text{q}}_{\text{B}}\text{=-3μC}\\ {\text{=-3×10}}^{\text{-6}}\text{\hspace{0.17em}}\text{C}\\ \text{Distance between two charges},\text{AB}=\text{2}0\text{cm}\\ \therefore \text{AO}=\text{OB}\\ =\text{1}0\text{cm}\\ \text{Net electric field at O}=\text{E}\\ \text{Electric field at point O caused by}+\text{3}\mu \text{C charge}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\\ \text{given}\text{\hspace{0.17em}}\text{as:}\\ {\text{E}}_{\text{1}}=\frac{{q}_{A}}{4\pi {\epsilon }_{0}{\left(AO\right)}^{2}}\text{}\\ \text{=}\frac{3×{10}^{-6}}{4\pi {\epsilon }_{0}{\left(\text{1}0×{10}^{-2}\right)}^{2}}\text{\hspace{0.17em}}N{C}^{-1}\text{\hspace{0.17em}}along\text{\hspace{0.17em}}OB\\ \text{Where},\text{\hspace{0.17em}}{\epsilon }_{0}=\text{Permittivity of free space}\\ \frac{1}{4\pi {\epsilon }_{0}}=9×{10}^{9}\text{\hspace{0.17em}}N{m}^{2}{C}^{-2}\\ \text{Magnitude of electric field at point O caused by}-\text{3}\mu \text{C}\\ \text{charge}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\\ {\text{E}}_{2}\text{}=|\frac{{q}_{B}}{4\pi {\epsilon }_{0}{\left(OB\right)}^{2}}|\\ =|\frac{-3×{10}^{-6}}{4\pi {\epsilon }_{0}{\left(\text{1}0×{10}^{-2}\right)}^{2}}|\\ \text{=}\frac{3×{10}^{-6}}{4\pi {\epsilon }_{0}{\left(\text{1}0×{10}^{-2}\right)}^{2}}\text{\hspace{0.17em}}N{C}^{-1}\text{\hspace{0.17em}}along\text{\hspace{0.17em}}OB\\ \therefore E={\text{E}}_{\text{1}}+{\text{E}}_{\text{2}}\\ =\frac{3×{10}^{-6}}{4\pi {\epsilon }_{0}{\left(\text{1}0×{10}^{-2}\right)}^{2}}+\frac{3×{10}^{-6}}{4\pi {\epsilon }_{0}{\left(\text{1}0×{10}^{-2}\right)}^{2}}\\ =2×\left[\left(9×{10}^{9}\right)×\frac{3×{10}^{-6}}{{\left(\text{1}0×{10}^{-2}\right)}^{2}}\right]\\ =5.4×{10}^{6}\text{\hspace{0.17em}}N{C}^{-1}\text{\hspace{0.17em}}\text{along OB}\\ {\text{E}}_{\text{2}}\text{}=\text{5}.\text{4}×\text{1}{0}^{\text{6}}{\text{NC}}^{\text{-1}}\text{along OB}\\ \therefore \text{The}\text{\hspace{0.17em}}\text{electric field at mid}-\text{point O is 5}.\text{4}×\text{1}{0}^{\text{6}}{\text{NC}}^{-\text{1}}\text{}\\ \text{along OB}.\\ \left(\text{b}\right)\text{A test charge of magnitude 1}.\text{5}×\text{1}{0}^{-\text{9}}\text{C is placed}\\ \text{at}\text{\hspace{0.17em}}\text{mid}-\text{point O}.\\ \text{Here,}\text{\hspace{0.17em}}\text{q}=\text{1}.\text{5}×\text{1}{0}^{-\text{9}}\text{C}\\ \text{Let}\text{\hspace{0.17em}}f\text{orce experienced by the test charge}=\text{F}\\ \therefore \text{F}=\text{qE}\\ \text{= 1}{\text{.5×10}}^{\text{-9}}\text{×5}{\text{.4×10}}^{\text{6}}\text{= 8}{\text{.1×10}}^{\text{-3}}\text{N;}\text{\hspace{0.17em}}\text{along BA}\text{.}\end{array}$

Q.9 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. $\begin{array}{l}{\text{The charges q}}_{\text{A}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{\text{q}}_{\text{B}}\text{\hspace{0.17em}}\text{can be located in a coordinate}\\ \text{frame of reference as shown in the given figure}\text{.}\\ \text{Magnitude of charge}\text{\hspace{0.17em}}\text{at}\text{\hspace{0.17em}}{\text{A, q}}_{\text{A}}\text{= 2}{\text{.5 × 10}}^{\text{-7}}\text{C}\\ \text{Magnitude of charge}\text{\hspace{0.17em}}\text{at}\text{\hspace{0.17em}}{\text{B, q}}_{\text{B}}\text{= -2}{\text{.5 × 10}}^{\text{-7}}\text{C}\\ \text{Total charge of the system,}\text{\hspace{0.17em}}{\text{q = q}}_{\text{A}}{\text{+ q}}_{\text{B}}\\ \text{= 2}{\text{.5 × 10}}^{\text{7}}\text{C – 2}{\text{.5 × 10}}^{\text{-7}}\text{C}\\ \text{= 0}\text{\hspace{0.17em}}\text{C}\\ \text{Distance between the}\text{\hspace{0.17em}}\text{two charges,}\text{\hspace{0.17em}}\text{d=15+15}\\ \text{=30 cm}\\ \text{=0}\text{.3 m}\\ \text{Electric dipole moment of the system is given as:}\\ \text{p}={\text{q}}_{\text{A}}×\text{d}\\ ={\text{q}}_{\text{B}}×\text{d}\\ =\text{2}.\text{5}×\text{1}{0}^{-\text{7}}×0.\text{3}\\ =\text{7}.\text{5}×\text{1}{0}^{-\text{8}}\text{C m along positive z}-\text{axis}\\ \therefore \text{Electric dipole moment of the system is}\text{\hspace{0.17em}}\text{7}.\text{5}×\text{1}{0}^{-\text{8}}\text{Cm}\\ \text{along positive}\text{\hspace{0.17em}}\text{z}-\text{axis}.\end{array}$

Q.10 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.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}{\text{electric dipole moment, p = 4×10}}^{\text{-9}}\text{\hspace{0.17em}}\text{Cm}\\ \text{Angle made by p with the}\text{\hspace{0.17em}}\text{uniform electric field, θ=30°}\\ \text{Magnitude}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}{\text{electric field, E = 5×10}}^{\text{4}}{\text{NC}}^{\text{-1}}\\ \text{Torque acting on the dipole is given by,}\\ \text{τ = pE sinθ}\\ {\text{=4×10}}^{\text{-9}}\text{×}\text{\hspace{0.17em}}{\text{5×10}}^{\text{4}}\text{sin30°}\\ {\text{=20×10}}^{\text{-5}}\text{×}\frac{\text{1}}{\text{2}}\\ {\text{=10}}^{\text{-4}}\text{\hspace{0.17em}}\text{Nm}\\ \text{The}\text{\hspace{0.17em}}\text{magnitude of the torque acting on the dipole is}\\ {\text{10}}^{\text{-4}}\text{\hspace{0.17em}}\text{Nm}\text{.}\end{array}$

Q.11 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.

$\begin{array}{l}\left(\text{a}\right)\text{\hspace{0.17em}}\text{Here, charge on the polythene piece,}\\ {\text{q=-3×10}}^{\text{-7}}\text{\hspace{0.17em}}\text{C}\\ \text{Charge on an electron, e = -1}{\text{.6×10}}^{\text{-19}}\text{C}\\ \text{Let}\text{\hspace{0.17em}}\text{number of electrons transferred to polythene}\\ \text{piece}\text{\hspace{0.17em}}\text{from}\text{\hspace{0.17em}}\text{wool= n}\\ \text{We}\text{\hspace{0.17em}}\text{can}\text{\hspace{0.17em}}\text{calculate}\text{\hspace{0.17em}}\text{n}\text{\hspace{0.17em}}\text{by using the relation,}\\ \text{q = ne}\\ \text{n=}\frac{\text{q}}{\text{e}}\text{=}\frac{{\text{-3×10}}^{\text{-7}}}{\text{-1}{\text{.6×10}}^{\text{-19}}}\\ \text{= 1}{\text{.87×10}}^{\text{12}}\\ \text{Number of electrons transferred from wool to}\\ \text{polythene is 1}{\text{.87 ×10}}^{\text{12}}\text{.}\\ \left(\text{b}\right)\text{Yes,}\text{\hspace{0.17em}}\text{transfer of mass takes place}\text{\hspace{0.17em}}\text{from}\text{\hspace{0.17em}}\text{wool}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\\ \text{the}\text{\hspace{0.17em}}\text{polythene}\text{\hspace{0.17em}}\text{piece}\text{.}\\ \text{As}\text{\hspace{0.17em}}\text{mass}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{electron,}\text{\hspace{0.17em}}{\text{m}}_{\text{e}}\text{= 9}{\text{.1×10}}^{\text{-31}}\text{kg}\\ \text{Total mass transferred from}\text{\hspace{0.17em}}\text{wool}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{polythene,}\text{\hspace{0.17em}}\text{m}\\ {\text{= m}}_{\text{e}}\text{× n}\\ \text{= 9}{\text{.1×10}}^{\text{-31}}\text{×1}{\text{.85×10}}^{\text{12}}\\ \text{= 1}{\text{.706×10}}^{\text{-18}}\text{kg}\\ \therefore A\text{negligible amount of mass is transferred from}\\ \text{wool to polythene}.\end{array}$

Q.12 (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.

$\begin{array}{l}\left(\text{a}\right)\text{Here,}\text{\hspace{0.17em}}{\text{charge on sphere A, q}}_{\text{A}}\text{=6}{\text{.5×10}}^{\text{-7}}\text{C}\\ {\text{Charge on sphere B, q}}_{\text{B}}\text{=6}{\text{.5×10}}^{\text{-7}}\text{C}\\ \text{Distance between spheres, r=50 cm}\\ \text{=0}\text{.5 m}\\ \text{Force of electrostatic}\text{\hspace{0.17em}}\text{repulsion between the two}\\ \text{spheres}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\\ \text{F=}\frac{{\text{q}}_{\text{A}}{\text{q}}_{\text{B}}}{4\pi {\epsilon }_{0}{r}^{2}}\\ \text{Here},\text{\hspace{0.17em}}{\epsilon }_{0}\text{}=P\text{ermittivity}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}f\text{ree space}=\text{9}×\text{1}{0}^{\text{9}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\\ \therefore \text{F=}\frac{\text{9}×\text{1}{0}^{\text{9}}×{\left(\text{6}.\text{5}×\text{1}{0}^{-\text{7}}\right)}^{2}}{{\left(0.\text{5}\right)}^{2}}\\ \text{= 1}{\text{.52 × 10}}^{\text{-2}}\text{N}\\ \therefore \text{The}\text{\hspace{0.17em}}\text{force between the two spheres is 1}.\text{52}×\text{1}{0}^{-\text{2}}\text{N}.\\ \left(\text{b}\right)\text{After doubling the amount}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{charge},\text{}\\ {\text{Charge on sphere A, q}}_{\text{A}}\text{= Charge on sphere B,}\text{\hspace{0.17em}}{\text{q}}_{\text{B}}\text{}\\ =\text{2}×\text{6}.\text{5}×\text{1}{0}^{-\text{7}}\text{C}\\ =\text{1}.\text{3}×\text{1}{0}^{-\text{6}}\text{C}\\ \text{As}\text{\hspace{0.17em}}\text{distance between the spheres is halved,}\\ \therefore r=\frac{0.\text{5}}{2}\\ =0.25\text{\hspace{0.17em}}m\\ \text{Force of repulsion between the two spheres}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as},\\ \text{F=}\frac{{\text{q}}_{\text{A}}{\text{q}}_{\text{B}}}{4\pi {\epsilon }_{0}{r}^{2}}\\ \text{=}\frac{{\text{9×10}}^{\text{9}}\text{×1}{\text{.3×10}}^{\text{-6}}\text{×1}{\text{.3×10}}^{\text{-6}}}{{\left(\text{0}\text{.25}\right)}^{\text{2}}}\\ \text{= 16×1}{\text{.52×10}}^{\text{-2}}\\ \text{= 0}\text{.243 N}\\ \therefore F\text{orce between the two spheres is}0.\text{243 N}.\end{array}$

Q.13 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.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}\text{initial charge on each sphere,}\\ \text{q = 6}{\text{.5 × 10}}^{\text{-7}}\text{C}\\ \text{Distance between the spheres A and B, r = 0}\text{.5 m}\\ \text{When sphere A is touched with an uncharged sphere C},\text{}\\ their\text{\hspace{0.17em}}ch\mathrm{arg}es\text{\hspace{0.17em}}are\text{\hspace{0.17em}}equally\text{\hspace{0.17em}}shared.\\ \therefore \text{Charge}\text{\hspace{0.17em}}\text{left}\text{\hspace{0.17em}}\text{on}\text{\hspace{0.17em}}\text{sphere}\text{\hspace{0.17em}}\text{A,}\text{\hspace{0.17em}}{\text{q}}_{\text{1}}\text{=}\frac{\text{6}{\text{.5 × 10}}^{\text{-7}}}{\text{2}}\\ =3.25×\text{1}{0}^{-\text{7}}\text{\hspace{0.17em}}C\\ \text{When sphere C with charge is brought in contact with}\\ \text{sphere B with charge q},\text{\hspace{0.17em}}their\text{\hspace{0.17em}}ch\mathrm{arg}es\text{\hspace{0.17em}}are\text{\hspace{0.17em}}shared\text{\hspace{0.17em}}\\ equally.\\ \therefore \text{Charge}\text{\hspace{0.17em}}\text{left}\text{\hspace{0.17em}}\text{on}\text{\hspace{0.17em}}\text{sphere}\text{\hspace{0.17em}}\text{B,}\text{\hspace{0.17em}}{\text{q}}_{\text{2}}\text{=}\frac{\left(\text{6}\text{.5+3}\text{.25}\right){\text{×10}}^{\text{-7}}}{\text{2}}\text{C}\\ =4.875×\text{1}{0}^{-\text{7}}\text{\hspace{0.17em}}C\\ As\text{\hspace{0.17em}}force\text{\hspace{0.17em}}of\text{\hspace{0.17em}}repulsion\text{\hspace{0.17em}}is\text{\hspace{0.17em}}given\text{\hspace{0.17em}}by\text{\hspace{0.17em}}the\text{\hspace{0.17em}}relation,\\ F=\frac{1}{4\pi {\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}\\ Here,\text{\hspace{0.17em}}\frac{1}{4\pi {\epsilon }_{0}}\text{\hspace{0.17em}}=9×{10}^{9}\text{\hspace{0.17em}}N{m}^{2}{C}^{-2}\\ \therefore \text{F=}\frac{{\text{9×10}}^{\text{9}}\text{×3}{\text{.25×10}}^{\text{-7}}\text{×4}{\text{.875×10}}^{\text{-7}}}{{\left(\text{0}\text{.5}\right)}^{\text{2}}}\\ \text{=5}{\text{.7×10}}^{\text{-3}}\text{\hspace{0.17em}}\text{N}\\ \therefore \text{The force of attraction between the two spheres is}\\ \text{5}.\text{7}0\text{3}×\text{1}{0}^{-\text{3}}\text{N}.\end{array}$

Q.14 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.15

Ans.

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

Q.16 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.17 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.

$\begin{array}{l}\left(\text{a}\right)\text{Here,}\text{\hspace{0.17em}}n\text{et outward flux through the surface of the}\\ \text{box},\text{}\varphi =\text{8}.0×\text{1}{0}^{\text{3}}{\text{N m}}^{\text{2}}{\text{C}}^{\text{-1}}\\ \text{For a body containing net charge q},\text{flux is given}\text{\hspace{0.17em}}\text{as:}\\ \varphi =\frac{q}{{\epsilon }_{0}}\\ Here,\text{\hspace{0.17em}}{\epsilon }_{0}=\text{Permittivity of free space}\\ =\text{8}.\text{854}×\text{1}{0}^{-\text{12}}{\text{N}}^{-\text{1}}{\text{C}}^{\text{2}}{\text{m}}^{-\text{2}}\\ \text{q}={\epsilon }_{0}\varphi \\ =\text{8}.\text{854}×\text{1}{0}^{-\text{12}}×\text{8}.0×\text{1}{0}^{\text{3}}\\ =\text{7}.0\text{8}×\text{1}{0}^{-\text{8}}\\ =0.0\text{7}\mu \text{C}\\ \therefore \text{Net charge inside the box is}0.0\text{7}\mu \text{C}.\\ \left(\text{b}\right)\text{If net flux through}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{surface}\text{\hspace{0.17em}}\text{is zero},\text{then as per}\\ \text{gauss theorem it can be concluded that net charge inside}\\ \text{the body is zero}.\text{In}\text{\hspace{0.17em}}\text{such}\text{\hspace{0.17em}}\text{situation,}\text{\hspace{0.17em}}t\text{he body may have}\\ \text{equal amount of positive and negative charges}.\end{array}$

Q.18 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:

$\begin{array}{l}{\varphi }_{total}=\frac{q}{{\epsilon }_{0}}\\ \therefore \text{Electric flux through the square},\text{\hspace{0.17em}}\varphi =\frac{{\varphi }_{total}}{6}\\ =\frac{1}{6}\frac{q}{{\epsilon }_{0}}\\ \text{Here},\text{\hspace{0.17em}}{\epsilon }_{0}=\text{Permittivity of free space}\\ =\text{8}.\text{854}×\text{1}{0}^{-\text{12}}{\text{N}}^{-\text{1}}{\text{C}}^{\text{2}}{\text{m}}^{-\text{2}}\\ \text{Point}\text{\hspace{0.17em}}\text{charge,}\text{\hspace{0.17em}}\text{q}=\text{1}0\text{}\mu \text{C}\\ =\text{1}0×\text{1}{0}^{-\text{6}}\text{C}\\ \therefore {\varphi }_{E}=\frac{1}{6}×\frac{\text{1}0×\text{1}{0}^{-\text{6}}}{\text{8}.\text{854}×\text{1}{0}^{-\text{12}}}\\ \text{= 1}.\text{88}×\text{1}{0}^{\text{5}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{1}}\\ \therefore \text{Electric flux through the square is 1}.\text{88}×\text{1}{0}^{\text{5}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{1}}.\end{array}$

Q.19 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.

$\begin{array}{l}Here,\text{\hspace{0.17em}}\text{Net charge inside the cube},\text{\hspace{0.17em}}q=\text{2}.0\text{}\mu \text{C}\\ =\text{2}×\text{1}{0}^{-\text{6}}\text{C}\\ \text{Net electric flux}\left({\varphi }_{\text{net}}\right)\text{through the cubic surface is}\\ \text{given by}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{relation},\\ {\varphi }_{\text{net}}=\frac{q}{{\epsilon }_{0}}\\ \text{Here},{\epsilon }_{0}=\text{Permittivity of free space}\\ =\text{8}.\text{854}×\text{1}{0}^{-\text{12}}{\text{N}}^{-\text{1}}{\text{C}}^{\text{2}}{\text{m}}^{-\text{2}}\\ \therefore \text{\hspace{0.17em}}{\varphi }_{\text{net}}=\frac{\text{2}×\text{1}{0}^{-\text{6}}}{\text{8}.\text{854}×\text{1}{0}^{-\text{12}}}\\ =\text{2}.\text{26}×\text{1}{0}^{\text{5}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{1}}\\ \therefore \text{Net electric flux through the surface is}\\ \text{2}.\text{26}×\text{1}{0}^{\text{5}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{1}}.\end{array}$

Q.20 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.

$\begin{array}{l}Here,\text{\hspace{0.17em}}radius\text{\hspace{0.17em}}of\text{\hspace{0.17em}}the\text{\hspace{0.17em}}sphere,\text{\hspace{0.17em}}r=10\text{\hspace{0.17em}}cm\\ \text{Electric field}\text{\hspace{0.17em}}\text{intensity},\text{\hspace{0.17em}}E=-\text{1}.\text{5}×\text{1}{0}^{\text{3}}{\text{NC}}^{\text{-1}}\\ \left(Negative\text{\hspace{0.17em}}sign\text{\hspace{0.17em}}for\text{\hspace{0.17em}}inward\text{\hspace{0.17em}}field\right)\\ \text{Distance from the centre}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}the\text{\hspace{0.17em}}sphere,\text{\hspace{0.17em}}d=\text{2}0\text{cm}\\ =0.\text{2 m}\\ \text{Let}\text{\hspace{0.17em}}\text{net}\text{\hspace{0.17em}}\text{charge}\text{\hspace{0.17em}}\text{on}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{sphere=q}\\ \text{Electric field intensity}\text{\hspace{0.17em}}\text{is given by the relation},\\ \because \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{d > r}\\ E=\frac{q}{4\pi {\epsilon }_{0}{d}^{2}}\text{\hspace{0.17em}}\\ Here,\text{\hspace{0.17em}}{\epsilon }_{0}=\text{Permittivity of free space}\\ =\text{9}×\text{1}{0}^{\text{9}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\\ \therefore q=E×4\pi {\epsilon }_{0}{d}^{2}\\ =\frac{\text{-1}.\text{5}×\text{1}{0}^{\text{3}}×{\left(0.\text{2}\right)}^{2}}{\text{9}×\text{1}{0}^{\text{9}}}\\ =-\text{6}.\text{67}×\text{1}{0}^{\text{9}}\text{C}\\ =-\text{6}.\text{67 nC}\\ \therefore \text{Net charge on the sphere is -6}.\text{67 nC}\text{.}\end{array}$

Q.21 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.

$\begin{array}{l}\left(\text{a}\right)\text{Here,}\text{\hspace{0.17em}}e\text{lectric flux},\text{}\varphi =-\text{1}.0×\text{1}{0}^{\text{3}}{\text{N m}}^{\text{2}}{\text{C}}^{\text{-1}}\\ \text{Radius of the}\text{\hspace{0.17em}}\text{Gaussian surface},\text{\hspace{0.17em}}R=\text{1}0.0\text{cm}\\ \text{Electric flux passing through a surface depends on the}\\ \text{net charge enclosed inside a body}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}i\text{t does not}\\ \text{depend on the size of the body}.\text{}\\ \therefore \text{If the radius}\text{\hspace{0.17em}}\text{of the}\text{\hspace{0.17em}}\text{Gaussian surface is doubled},\text{then}\\ \text{the flux passing through the surface remains the}\text{\hspace{0.17em}}\text{same}\text{.}\\ \left(\text{b}\right)\text{Electric flux is given as:}\\ \varphi \text{=}\frac{q}{{\epsilon }_{0}}\\ \text{Here},\text{\hspace{0.17em}}\text{q}=\text{Net charge enclosed by the surface}\\ {\epsilon }_{0}\text{}=\text{Permittivity of free space}\\ =\text{8}.\text{854}×\text{1}{0}^{-\text{12}}\text{\hspace{0.17em}}{\text{N}}^{-\text{1}}{\text{C}}^{\text{2}}{\text{m}}^{-\text{2}}\\ \therefore q=\varphi {\epsilon }_{0}=-\text{1}.0×\text{1}{0}^{\text{3}}×\text{8}.\text{854}×\text{1}{0}^{-\text{12}}\\ =\text{}-\text{8}.\text{854}×\text{1}{0}^{-\text{9}}\text{C}=\text{}-\text{8}.\text{854 nC}\\ \therefore \text{The value of the point charge is}-\text{8}.\text{854 nC}.\end{array}$

Q.22 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.

$\begin{array}{l}\left(\text{a}\right)\text{Here,}\text{\hspace{0.17em}}d\text{iameter of sphere},\text{d}=\text{2}.\text{4 m}\\ \therefore \text{Radius of sphere},\text{r}=\frac{\text{d}}{2}\\ \text{= 1}.\text{2 m}\\ \text{Surface charge density}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{sphere},\text{}\sigma =\text{8}0.0\text{}\mu {\text{Cm}}^{\text{-2}}\text{}\\ =\text{8}0×\text{1}{0}^{-\text{6}}{\text{Cm}}^{\text{-2}}\\ \text{Total charge on the surface of the sphere}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\\ \text{Q}=\sigma ×4\pi {r}^{2}\\ =\text{8}0×\text{1}{0}^{-\text{6}}×\text{4}×\text{3}.\text{14}×{\left(\text{1}.\text{2}\right)}^{\text{2}}\\ =\text{1}.\text{447}×\text{1}{0}^{-\text{3}}\text{C}\\ \therefore \text{Charge on the sphere is 1}.\text{447}×\text{1}{0}^{-\text{3}}\text{C}.\\ \left(\text{b}\right)\text{}Here,\text{\hspace{0.17em}}Q=\text{1}.\text{447}×\text{1}{0}^{-\text{3}}\text{C}\\ \text{Total electric flux leaving the surface of a sphere is}\\ \text{given by the relation},\\ {\varphi }_{total}=\frac{\text{Q}}{{\epsilon }_{0}}\\ \text{Here},\text{\hspace{0.17em}}{\epsilon }_{0}=\text{Permittivity of free space}\\ =\text{8}.\text{854}×\text{1}{0}^{-\text{12}}{\text{N}}^{-\text{1}}{\text{C}}^{\text{2}}{\text{m}}^{-\text{2}}\\ \therefore {\varphi }_{total}=\frac{\text{1}.\text{447}×\text{1}{0}^{-\text{3}}}{\text{8}.\text{854}×\text{1}{0}^{-\text{12}}}\\ =\text{1}.\text{63}×\text{1}{0}^{\text{8}}{\text{NC}}^{-\text{1}}{\text{m}}^{\text{2}}\\ \therefore \text{The total electric flux leaving the surface of the}\\ \text{sphere is 1}.\text{63}×\text{1}{0}^{\text{8}}{\text{NC}}^{-\text{1}}{\text{m}}^{\text{2}}.\end{array}$

Q.23 An infinite line charge produces a field of 9 × 104 NC-1 at a distance of 2 cm. Calculate the linear charge density.

Ans.

$\begin{array}{l}\\ Here,\text{\hspace{0.17em}}\\ \text{Electric field},\\ E=\text{9}×\text{1}{0}^{\text{4}}{\text{NC}}^{\text{-1}}\\ Dis\mathrm{tan}ce,\text{\hspace{0.17em}}\\ d=2\text{\hspace{0.17em}}cm\\ =0.02\text{\hspace{0.17em}}m\\ \text{Electric field produced by the infinite line charges at a}\\ \text{distance d having linear charge density}\lambda \text{is given as:}\\ \text{E=}\frac{\lambda }{2\pi {\epsilon }_{0}d}\\ \therefore \lambda =2\pi {\epsilon }_{0}d×\text{E}\\ Here,\\ {\epsilon }_{0}=\text{Permittivity of free space}\\ \frac{1}{4\pi {\epsilon }_{0}}=\text{9}×\text{1}{0}^{\text{9}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\\ \therefore \lambda =\frac{0.02×\text{9}×\text{1}{0}^{\text{4}}}{2×\text{9}×\text{1}{0}^{\text{9}}}\\ =\text{1}{0}^{-7}{\text{Cm}}^{\text{-1}}\\ \therefore \text{The linear charge density is}0.1\text{}\mu {\text{Cm}}^{\text{-1}}.\end{array}$

Q.24

$\begin{array}{l}\text{Two large, thin metal plates are parallel and close to each other}\text{. On their inner faces,}\\ \text{the plates have surface charge densities of opposite signs and of magnitude}\\ \text{17}{\text{.0×10}}^{\text{-22}}\text{\hspace{0.17em}}{\text{Cm}}^{\text{-2}}\text{. What is}\stackrel{\to }{E}\text{:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{a}\right)\text{in the}\text{\hspace{0.17em}}\text{outer region of the first plate,}\left(\text{b}\right)\text{in the outer}\\ \text{region of the second plate, and}\left(\text{c}\right)\text{between the plates?}\end{array}$

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.

$Here, surface charge density, σ = 17 .0×10 -22 Cm -2 Since charge is not enclosed by the respective plates in the regions, I and III, ∴Electric field, E = 0, in regions I and III. Thus, ( a ) and ( b ) have net electric field zero. ( c )Electric field E in region II is given by, E = σ ε 0 Here, ε 0 =8 .854×10 -12 N -1 C 2 m -2 ∴E = 17 .0×10 -22 Cm -2 8 .854×10 -12 N -1 C 2 m -2 = 1 .92×10 -10 NC -1 ∴Electric field between the plates is 1 .92×10 -10 NC -1 .$

Q.25 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.

$\begin{array}{l}\text{Here,\hspace{0.17em}}\mathrm{e}\text{xcess electrons on an oil drop},\text{n = 12}\\ \text{Magnitude\hspace{0.17em}of\hspace{0.17em}Electric field},{\text{E = 2.55×10}}^{\text{4}}{\text{NC}}^{\text{-1}}\\ \text{Density of oil},{\text{ρ = 1.26 gmcm}}^{\text{-3}}\text{}\\ {\text{= 1.26 × 10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\\ {\text{Acceleration due to gravity, g = 9.81 m s}}^{\text{-2}}\\ {\text{Charge on an electron, e = 1.6×10}}^{\text{-19}}\text{C}\\ \text{Let\hspace{0.17em}radius of\hspace{0.17em}oil drop = r}\\ \text{As\hspace{0.17em}the\hspace{0.17em}oil\hspace{0.17em}drop\hspace{0.17em}is\hspace{0.17em}stationary,}\\ \therefore \text{Force due to electric field = Weight of the oil drop}\\ \text{F = W}\\ \text{As\hspace{0.17em}F=Eq\hspace{0.17em}and\hspace{0.17em}W=mg}\\ \therefore \text{Eq = mg}\to \text{(i)}\mathrm{}\\ \text{Charge on oil drop,\hspace{0.17em}q= ne→(ii)}\\ \text{Mass of oil drop,\hspace{0.17em}m}\\ \text{= Volume of the oil drop × Density of oil}\\ \text{=}\frac{4}{3}{\mathrm{\pi r}}^{3}×\mathrm{\rho }\to \left(\mathrm{iii}\right)\\ \mathrm{Substituting}\text{\hspace{0.17em}}\mathrm{equations}\text{\hspace{0.17em}(ii)\hspace{0.17em}and\hspace{0.17em}}\left(\mathrm{iii}\right),\text{\hspace{0.17em}}\mathrm{in}\text{\hspace{0.17em}}\mathrm{equation}\text{\hspace{0.17em}}\left(\mathrm{i}\right),\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\\ \mathrm{get}\\ \text{Ene=}\frac{4}{3}{\mathrm{\pi r}}^{3}×\mathrm{\rho g}\\ \therefore \mathrm{r}=\sqrt{\frac{3\text{Ene}}{4\mathrm{\pi \rho g}}}\\ \text{=}\sqrt[\text{3}]{\frac{{\text{3×2.55×10}}^{\text{4}}{\text{×12×1.6×10}}^{\text{-19}}}{{\text{4×3.14×1.26×10}}^{\text{3}}\text{×9.81}}}\\ \text{=}\sqrt[\text{3}]{{\text{946.09×10}}^{\text{-21}}}\\ {\text{=9.82×10}}^{\text{-7}}\text{\hspace{0.17em}m}\\ {\text{= 9.82×10}}^{\text{-4}}\text{mm}\\ \therefore \text{The radius of the oil drop is 9}.\text{82}×\text{1}{0}^{-\text{4}}\text{mm}.\end{array}$

Q.26 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.27 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 ,

$\begin{array}{l}\frac{dE}{dl}={10}^{5}\text{\hspace{0.17em}}{\text{NC}}^{\text{-1}}\\ \text{Force experienced by the system is given as:}\\ \text{F = qE}\\ \text{=q}\frac{\text{dE}}{\text{dl}}\text{×dl}\\ \text{=p×}\frac{\text{dE}}{\text{dl}}\\ {\text{=-10}}^{\text{-7}}{\text{×10}}^{\text{5}}{\text{= -10}}^{\text{-2}}\text{\hspace{0.17em}}\text{N}\end{array}$

$\begin{array}{l}\text{The force}\text{\hspace{0.17em}}on\text{\hspace{0.17em}}the\text{\hspace{0.17em}}dipole\text{is}-\text{1}{0}^{-\text{2}}\text{N in the negative z}-\\ \text{direction}\\ \therefore \text{Angle between electric field and dipole moment,}\text{\hspace{0.17em}}\theta \\ \text{=18}0°.\\ \text{Torque is given as:}\text{\hspace{0.17em}}\tau \text{=pE sinθ= pE sin180°= 0}\\ \therefore \text{Torque experienced by the system is zero}.\end{array}$

Q.28 (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.

$\begin{array}{l}\text{Let charge inside the conductor}=\text{q \hspace{0.17em}}\\ \text{According to Gauss}’\text{s law},\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{have}\\ \text{Flux},\text{\hspace{0.17em}}\mathrm{\varphi }=\stackrel{\to }{\mathrm{E}}.\stackrel{\to }{\mathrm{ds}}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_{0}}\\ \text{Here,\hspace{0.17em}}{\mathrm{\epsilon }}_{0}=\text{Permittivity of free space}\\ \text{As\hspace{0.17em}net\hspace{0.17em}electric\hspace{0.17em}field\hspace{0.17em}inside\hspace{0.17em}a\hspace{0.17em}charged\hspace{0.17em}conductor},\text{E}=0\\ \therefore \frac{\mathrm{q}}{{\mathrm{\epsilon }}_{0}}=0\\ \mathrm{As}\text{\hspace{0.17em}}{\mathrm{\epsilon }}_{0}\ne 0\\ \therefore \mathrm{q}=0\\ \therefore \mathrm{Charge}\text{\hspace{0.17em}}\mathrm{inside}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{conductor}=0.\\ \therefore \mathrm{Entire}\text{\hspace{0.17em}}\mathrm{charge}\text{\hspace{0.17em}}\mathrm{Q}\text{\hspace{0.17em}}\mathrm{on}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{conductor}\text{\hspace{0.17em}}\mathrm{appears}\text{\hspace{0.17em}}\mathrm{on}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\\ \mathrm{outer}\text{\hspace{0.17em}}\mathrm{surface}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{conductor}.\end{array}$

(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.29 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.

$\begin{array}{l}\text{Consider a conductor with a tiny}\text{\hspace{0.17em}}\text{hole}\text{\hspace{0.17em}}\text{into}\text{\hspace{0.17em}}\text{its}\text{\hspace{0.17em}}\text{surface}.\text{}\\ \text{Electric field inside the hole =0}\\ \text{Suppose E}\text{\hspace{0.17em}}is\text{\hspace{0.17em}}\text{electric field outside the conductor,}\text{\hspace{0.17em}}\text{q}\text{\hspace{0.17em}}is\text{}\\ \text{the}\text{\hspace{0.17em}}\text{electric}\text{\hspace{0.17em}}\text{charge}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\sigma \text{\hspace{0.17em}}is\text{\hspace{0.17em}}the\text{\hspace{0.17em}}surface\text{\hspace{0.17em}}ch\mathrm{arg}e\text{\hspace{0.17em}}density.\\ Electric\text{\hspace{0.17em}}ch\mathrm{arg}e,\text{\hspace{0.17em}}q=\sigma ×ds\\ \text{According to Gauss}’\text{s law},\text{\hspace{0.17em}}we\text{\hspace{0.17em}}have\\ \oint \stackrel{\to }{E}.\stackrel{\to }{ds}=\frac{q}{{\epsilon }_{\text{0}}}\\ Here,\text{\hspace{0.17em}}{\epsilon }_{\text{0}}=Permittivity\text{\hspace{0.17em}}of\text{\hspace{0.17em}}free\text{\hspace{0.17em}}space\\ \therefore Eds=\frac{\sigma ×ds}{{\epsilon }_{\text{0}}}\\ E=\frac{\sigma }{{\epsilon }_{\text{0}}}\\ or\text{\hspace{0.17em}}\stackrel{\to }{E}=\frac{\sigma }{{\epsilon }_{\text{0}}}\stackrel{^}{n}\\ \therefore \text{Electric field just outside the conductor is}\frac{\sigma }{{\epsilon }_{\text{0}}}\stackrel{^}{n}.\text{}\\ \text{This field is superposition of field due to filled}\text{\hspace{0.17em}}\text{up}\text{\hspace{0.17em}}\text{hole}\\ \text{and the field due to the rest of the}\text{\hspace{0.17em}}\text{charged conductor}.\text{}\\ \text{These two}\text{\hspace{0.17em}}\text{fields are equal and opposite inside the}\\ \text{conductor}\text{.}\text{\hspace{0.17em}}\text{Therefore,}\text{\hspace{0.17em}}\text{there}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}no\text{\hspace{0.17em}}electric\text{\hspace{0.17em}}field\text{\hspace{0.17em}}inside\text{\hspace{0.17em}}\\ the\text{\hspace{0.17em}}conductor.\text{\hspace{0.17em}}However,\text{\hspace{0.17em}}these\text{\hspace{0.17em}}fields\text{\hspace{0.17em}}are\text{\hspace{0.17em}}\text{equal in}\\ \text{magnitude and direction outside the conductor}.\\ \therefore \text{The electric}\text{\hspace{0.17em}}\text{field}\text{\hspace{0.17em}}\text{at}\text{\hspace{0.17em}}P,\text{due to each}\text{\hspace{0.17em}}\text{part}\text{\hspace{0.17em}}\text{=}\frac{1}{2}\text{\hspace{0.17em}}\stackrel{\to }{E}\text{}\\ \text{=}\frac{\sigma }{2{\epsilon }_{\text{0}}}\stackrel{^}{n}\end{array}$

Q.30 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.

$\begin{array}{l}\text{Let E be the electric field at point S because of wire AB}.\\ \text{Consider a small element of length dx on the wire with}\\ \text{OZ}=\text{x}\\ \text{Let q be the charge on this element}.\\ \therefore \text{q}=\lambda dx\\ \text{Electric field due to the element},\\ dE=\frac{1}{4\pi {\epsilon }_{0}}\frac{\lambda dx}{{\left(SZ\right)}^{2}}\\ As,\text{\hspace{0.17em}}SZ=\sqrt{{l}^{2}+{x}^{2}}\\ \therefore dE=\frac{\lambda dx}{4\pi {\epsilon }_{0}\left({l}^{2}+{x}^{2}\right)}\\ The\text{\hspace{0.17em}}electric\text{\hspace{0.17em}}field\text{\hspace{0.17em}}can\text{\hspace{0.17em}}be\text{\hspace{0.17em}}resolved\text{\hspace{0.17em}}\mathrm{int}o\text{\hspace{0.17em}}two\text{\hspace{0.17em}}rec\mathrm{tan}gular\text{\hspace{0.17em}}\\ components:\text{\hspace{0.17em}}\text{dE}\mathrm{cos}\theta \text{\hspace{0.17em}}perpendicular\text{\hspace{0.17em}}to\text{\hspace{0.17em}}AB\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{dE}\mathrm{sin}\theta \text{\hspace{0.17em}}\\ parallel\text{\hspace{0.17em}}to\text{\hspace{0.17em}}AB.\\ \text{Only the perpendicular component dEcos}\theta \text{\hspace{0.17em}}\text{affects}\\ \text{point S}.\\ \therefore \text{Effective component}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{electric field at point S due to}\\ \text{the element dx,}\text{\hspace{0.17em}}{\text{dE}}_{\text{1}}=\text{dE}\mathrm{cos}\theta \\ \therefore {\text{dE}}_{\text{1}}=\frac{\lambda \text{dx}\mathrm{cos}\theta }{4\pi {\epsilon }_{0}\left({l}^{2}+{x}^{2}\right)}\to \left(i\right)\\ In\text{\hspace{0.17em}}\Delta SZO,\\ \mathrm{tan}\theta =\frac{x}{l}\\ \therefore x=l\mathrm{tan}\theta \to \left(ii\right)\\ \text{Differentiating equation}\left(ii\right),\text{we get}\\ \frac{dx}{d\theta }=l{\mathrm{sec}}^{2}\theta \\ \therefore dx=l{\mathrm{sec}}^{2}\theta d\theta \to \left(iii\right)\\ \text{From equation}\left(ii\right),\\ {x}^{2}+{l}^{2}={l}^{2}+{l}^{2}{\mathrm{tan}}^{2}\theta \\ \therefore {l}^{2}\left(1+{\mathrm{tan}}^{2}\theta \right)={l}^{2}{\mathrm{sec}}^{2}\theta \\ {x}^{2}+{l}^{2}={l}^{2}{\mathrm{sec}}^{2}\theta \to \left(iv\right)\\ \text{Substituting equations}\left(iii\right)\text{and}\left(iv\right)\text{in equation}\left(i\right),\text{we}\\ get\\ d{E}_{1}=\frac{\lambda l{\mathrm{sec}}^{2}\theta d\theta }{4\pi {\epsilon }_{0}{l}^{2}{\mathrm{sec}}^{2}\theta }×\mathrm{cos}\theta \\ \therefore d{E}_{1}=\frac{\lambda \mathrm{cos}\theta d\theta }{4\pi {\epsilon }_{0}l}\to \left(v\right)\\ \text{The wire is so long that angle}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\text{tends from –}\frac{\pi }{2}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}+\frac{\pi }{2}.\\ \text{Integrating equation}\left(\text{5}\right),{\text{we obtain the value of E}}_{\text{1}}\text{as},\\ \underset{\text{–}\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}d{E}_{1}=\underset{\text{–}\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{\lambda \mathrm{cos}\theta d\theta }{4\pi {\epsilon }_{0}l}\\ {E}_{1}=\frac{\lambda }{4\pi {\epsilon }_{0}l}{\left[\mathrm{sin}\theta \right]}_{\text{–}\frac{\pi }{2}}^{\text{\hspace{0.17em}}\frac{\pi }{2}}\\ =\frac{\lambda }{4\pi {\epsilon }_{0}l}×2\\ \therefore {E}_{1}=\frac{\lambda }{2\pi {\epsilon }_{0}l}\\ \therefore \text{The}\text{\hspace{0.17em}}\text{electric field due to long wire is}\frac{\lambda }{2\pi {\epsilon }_{0}l}.\end{array}$

Q.31 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.

$\begin{array}{l}\text{Here,\hspace{0.17em}}\mathrm{t}\text{otal\hspace{0.17em}number\hspace{0.17em}of\hspace{0.17em}quarks\hspace{0.17em}in\hspace{0.17em}a\hspace{0.17em}proton=3}\\ \text{Let number\hspace{0.17em}of up quarks in a proton}=\text{n}\\ \therefore \text{Number of down quarks in a proton}=\text{3-n}\\ \text{Total\hspace{0.17em}charge\hspace{0.17em}on\hspace{0.17em}the\hspace{0.17em}proton\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:}\\ \left(\frac{2}{3}\mathrm{e}\right)\mathrm{n}+\left(-\frac{1}{3}\mathrm{e}\right)\left(\text{3-n}\right)=+\mathrm{e}\\ \frac{2}{3}\mathrm{en}-\mathrm{e}+\frac{\mathrm{ne}}{3}=\mathrm{e}\\ \\ \mathrm{ne}\left(\frac{2}{3}+\frac{1}{3}\right)=2\mathrm{e}\\ \therefore \mathrm{ne}=2\mathrm{e}\\ \therefore \mathrm{n}=2\\ \therefore \mathrm{Number}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{up}\text{\hspace{0.17em}}\mathrm{quarks}\text{\hspace{0.17em}}\mathrm{in}\text{\hspace{0.17em}}\mathrm{a}\text{\hspace{0.17em}}\mathrm{proton}=2\\ \mathrm{Number}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{down}\text{\hspace{0.17em}}\mathrm{quarks}\text{\hspace{0.17em}}\mathrm{in}\text{\hspace{0.17em}}\mathrm{a}\text{\hspace{0.17em}}\mathrm{proton}=3-2=1\\ \mathrm{A}\text{\hspace{0.17em}}\mathrm{proton}\text{\hspace{0.17em}}\mathrm{can}\text{\hspace{0.17em}}\mathrm{be}\text{\hspace{0.17em}}\mathrm{represented}\text{\hspace{0.17em}}\mathrm{as}:\mathrm{p}⇒\mathrm{UUd}\\ \text{Here,\hspace{0.17em}}\mathrm{t}\text{otal\hspace{0.17em}number\hspace{0.17em}of\hspace{0.17em}quarks\hspace{0.17em}in\hspace{0.17em}a\hspace{0.17em}proton=3}\\ \text{Let number\hspace{0.17em}of up quarks in a neutron}=\text{n}\\ \therefore \text{Number of down quarks in a neutron}=\text{3-n}\\ \text{Total\hspace{0.17em}charge\hspace{0.17em}on\hspace{0.17em}the\hspace{0.17em}neutron\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:}\\ \left(\frac{2}{3}\mathrm{e}\right)\mathrm{n}+\left(-\frac{1}{3}\mathrm{e}\right)\left(\text{3-n}\right)=0\\ \frac{2}{3}\mathrm{en}-\mathrm{e}+\frac{\mathrm{ne}}{3}=\mathrm{o}\\ \mathrm{ne}\left(\frac{2}{3}+\frac{1}{3}\right)=0\\ \therefore \mathrm{ne}=\mathrm{e}\\ \therefore \mathrm{n}=1\\ \therefore \mathrm{Number}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{up}\text{\hspace{0.17em}}\mathrm{quarks}\text{\hspace{0.17em}}\mathrm{in}\text{\hspace{0.17em}}\mathrm{a}\text{\hspace{0.17em}}\mathrm{neutron}=1\\ \mathrm{Number}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{down}\text{\hspace{0.17em}}\mathrm{quarks}\text{\hspace{0.17em}}\mathrm{in}\text{\hspace{0.17em}}\mathrm{a}\text{\hspace{0.17em}}\mathrm{neutron}=3-1=2\\ \mathrm{A}\text{\hspace{0.17em}}\mathrm{neutron}\text{\hspace{0.17em}}\mathrm{can}\text{\hspace{0.17em}}\mathrm{be}\text{\hspace{0.17em}}\mathrm{represented}\text{\hspace{0.17em}}\mathrm{as}:\mathrm{p}⇒\mathrm{Udd}\end{array}$

Q.32 (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.33 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.

$\begin{array}{l}\text{Here, }\\ \text{Charge on the particle of mass m}=-\text{q}\\ \text{Velocity of particle}={\text{v}}_{\text{x}}\\ \text{Length of plates}=\text{L}\\ \text{Magnitude of the uniform electric field}=\text{E}\\ \text{Mechanical force is given as: F}\\ =\text{Mass}\left(\text{m}\right)×\text{Acceleration}\left(\text{a}\right)\\ \therefore \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\\ \mathrm{Electric}\text{ }\mathrm{force}\text{ }\mathrm{is}\text{ }\mathrm{given}\text{ }\mathrm{as}:\text{ }\mathrm{F}=\mathrm{qE}\\ \therefore \text{Acceleration},\text{ }\\ \mathrm{a}=\frac{\mathrm{qE}}{\mathrm{m}}\to \left(\mathrm{i}\right)\\ \text{Time taken by the particle to cross the field is given as:}\\ \text{t=}\frac{\mathrm{distance}}{\mathrm{velocity}}\\ =\frac{\mathrm{length}\text{ }\mathrm{of}\text{ }\mathrm{plate}}{\mathrm{velocity}\text{ }\mathrm{of}\text{ }\mathrm{particle}}\\ =\frac{\mathrm{L}}{{\mathrm{v}}_{\mathrm{x}}}\to \left(\mathrm{ii}\right)\\ \text{Along vertical direction},\text{initial velocity},\\ \text{u}=0\\ \text{According to third equation of motion},\text{the vertical}\\ \text{deflection s of the particle can be obtained as},\\ \mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^{2}\\ \mathrm{s}=0+\frac{1}{2}\left(\frac{\mathrm{qE}}{\mathrm{m}}\right){\left(\frac{\mathrm{L}}{{\mathrm{v}}_{\mathrm{x}}}\right)}^{2}\\ \therefore \mathrm{s}=\frac{{\mathrm{qEL}}^{2}}{2{\mathrm{mv}}_{\mathrm{x}}^{2}}\to \left(\mathrm{iii}\right)\\ \therefore \text{Vertical deflection of the particle at the far edge of}\\ \text{the plate is }\frac{{\mathrm{qEL}}^{2}}{2{\mathrm{mv}}_{\mathrm{x}}^{2}}.\text{}\\ \text{This is exacly similar to the motion of horizontal}\\ \text{projectiles under the effect of gravity}.\end{array}$

Q.34 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.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}{\text{velocity of particle, v}}_{\text{x}}\text{= 2}{\text{.0×10}}^{\text{6}}{\text{ms}}^{\text{-1}}\\ \text{Separation of the plates, d = 0}\text{.5 cm = 0}\text{.005 m}\\ \text{Electric field between the plates, E=9}{\text{.1×10}}^{\text{2}}{\text{NC}}^{\text{-1}}\\ \text{Charge of electron, q = 1}{\text{.6 × 10}}^{\text{-19}}\text{C}\\ {\text{Mass of electron, m}}_{\text{e}}\text{= 9}{\text{.1 × 10}}^{\text{-31}}\text{kg}\\ \text{Let the electron strikes the upper plate at the end of}\\ \text{plate L,}\text{\hspace{0.17em}}\text{as}\text{\hspace{0.17em}}\text{soon}\text{\hspace{0.17em}}\text{as}\text{\hspace{0.17em}}\text{its deflection is d}\text{.}\\ \therefore d=\frac{{\text{qEL}}^{2}}{2m{v}_{x}{}^{2}}\\ \therefore L=\sqrt{\frac{2×0.005×9.1×{10}^{-31}×{\left(\text{2}.0×\text{1}{0}^{\text{6}}\right)}^{2}}{\text{1}.\text{6}×\text{1}{0}^{-\text{19}}×\text{9}.\text{1}×\text{1}{0}^{\text{2}}}}\\ =\sqrt{0.025×{10}^{-2}}\\ =\sqrt{2.5×{10}^{-4}}\\ =1.6×{10}^{-2}\text{\hspace{0.17em}}m\\ =1.6\text{\hspace{0.17em}}cm\\ \therefore \text{The}\text{\hspace{0.17em}}\text{electron will strike the upper plate after}\\ \text{travelling 1}.\text{6 cm}.\end{array}$