# Important Questions Class 12 Physics Chapter 2

## Important Questions for CBSE Class 12 Physics Chapter 2 – Electrostatic Potential and Capacitance

These Important Questions in Class 12 Physics Chapter 2 will teach students about electrostatic potential and capacitance. These Chapter 2 Class 12 Physics Important Questions will prepare students for their exams. After studying these Physics Class 12 chapter 2 Important Questions, students will be able to solve many CBSE sample papers.

### Study Important Questions for Class 12 Physics Chapter 2 – Electrostatic Potential and Capacitance

Electrostatic Potential and Capacitance is an important chapter and going through these important question notes can help you understand the exam pattern. These Class 12 Physics Chapter 2 Important Questions contain important formulas and CBSE extra questions that can help you test student’s understanding. In addition, these notes are crafted as per the prescribed CBSE syllabus and CBSE revision notes.

Q.1. Why does a dielectric’s internal electric field weaken when exposed to an external electric field?

Ans. The polarisation causes an internal electric field that is opposite to the external electric field inside a dielectric, which causes the net electric field to decrease when the dielectric is exposed to an external electric field.

Q.2. A 10 cm square with a 500 C charge in the centre. Find the amount of effort required to move a charge of 10C between two square spots that are diagonally opposed.

Ans. Since these two spots on the square will be equipotential, the work required to move a charge of 10C between them will be zero.

Q.3. When a 10C charge is present in the square’s centre, how much work is required to move a 2C point shift from corner A to corner B?

Ans. Points A and B are equally distant from point O. Hence the work done is equal to zero.

VA=VB

Therefore, the work done=0

Q.4. What physical effort is expended when an electric dipole’s equatorial axis is traversed by a test charge q over a distance of 1 cm?

Ans. Given that the equatorial axis potential is V = 0

∴W = qV = 0,

Q.5. A voltmeter connects the plates of a charged capacitor. What will happen when the capacitor’s plates are separated further by voltmeter reading?

Ans. Capacitance, area, distance, and dielectric constant are all related.

C=A0dC∝1d

Hence, the capacitance will decrease if the distance increases.

Since V=QC and there is a constant charge on the capacitor,

Therefore, the voltmeter reading will increase.

Q.6. A hollow metal sphere with a 10 cm radius is charged to a surface potential of 5 V. What potential exists in the sphere’s centre?

Ans. The potential in the centre of a hollow metal sphere will be 5 V because it functions as an equipotential surface.

Q.7. Why must every point of an empty charged conductor’s electrostatic potential be the same?

Ans. Since there is no electric field within the hollow-charged conductor, there is no effort expended in moving the test charge. The electrostatic potential is hence constant throughout a hollow-charged conductor.

Q.8. Imagine a parallel plate capacitor having air in between the plates has a capacitance of  8pF (1pF = 1012 F). What would the capacitance be if the space between the plates is halved and the space between them is taken over with a substance of dielectric constant 6?

Ans. For air, capacitance can be written as C0=A0d

C0 = 8pF=810-12F

Now, d’=d2 and k=6

C’=AC0d’2K

C’=810-1226

C’=9610-12pF

Q.9. If there are 1 and 2 volts of electric flux entering and exiting a closed surface, respectively, what is the surface’s internal electric charge?

Ans. Net flux can be represented as 2 -1

since  = q0

Q= (2 -1) 0

The electric charge that is present on the surface can be represented as

Q=0

Q.10. (i) Is it possible for two equipotential surfaces to intersect? Give reasons.

(ii) At positions A (0, 0, -a) and B (0, 0, +a), respectively, are two charges, -q and +q. In transferring a test charge from location P (7, 0, 0) to point Q (-3, 0, 0), how much work is involved?

Ans. (i) This is incorrect because if they connect, the electric field there would be pointing in two distinct directions. If they do, there will be two potential values present at the junction. As a result, two equipotential surfaces cannot intersect because this is not conceivable.

(ii) Since V = 0 at every point on the dipole’s equatorial line, which includes points P and Q, there will be no work done. Additionally, since any force of a charge is perpendicular to the equatorial line, no work is done.

Q.11. Each vertex of a regular hexagon having 10 cm of side length carries a charge of.

Calculate the potential in the hexagon’s centre.

Ans: The given illustration depicts six identical charges named q at the vertices of a regular hexagon.

In the case, Charge, q = 5C = 510-5 C

AB=BC=CD=DE=EF=FA = 10cm

Each vertex’s distance from the hexagon’s centre, O, d, is 10 cm, and the electric potential at point O is 10 cm.

V = 6q40d

Where

0 is the permittivity of free space,

140 = 9109NC-2m-2

V = 69109510-50.1

V = 2.7106V

Hence, 2.7106V is the potential at the centre of the hexagon.

Q.12. A spherical conductor with a radius of 12 cm has a 1.6 107 C charge evenly dispersed over its surface. What will the electric field be in the following cases?

• Inside the sphere
• Just outside the sphere
• At a location 18 cm from the sphere’s centre

Ans. (a) Radius of the spherical conductor can be represented as r = 12cm = 0.12m

The conductor is evenly covered with charge, which can be applied as,

q = 1.610-7C

There is no electric field inside of a spherical conductor. This is due to the fact that charges will flow to neutralise any fields present inside the conductor.

(b) The electric field E present just outside the conductor can be represented by the following relation,

E = q40r2

Where,

0 is the permittivity of free space,

140 = 9109NC-1m-2

E =1.610-7910-9(0.12)2

E = 105 NC-1

Hence, the electric field that will exist just outside the sphere will be 105 NC-1.

(c) The distance between the point from the centre, d = 18cm = 0.18m

E=q40d2

E= 91091.610-7(1810-2)

E = 4.4104 N/C

Hence, the electric field having an 18 cm distance from the centre of the sphere will be 4.4104 N/C.

Q.13. A constant electric field holds an electric dipole in place.

(i) Demonstrate that there is no net force operating on it.

(ii) The dipole is parallel to the field in alignment.

Find the work that was done to rotate it via a 180° angle.

Ans. (i) Force resulting from charge -q is -qE acting on point A

Point B is exposed to a force from charge +q. is + qE.

The net force acting on equals -qE plus qE, or 0. (zero)

Hence, the net force acting on an electric dipole held in a uniform electric field is zero.

(ii) W = -pE(cos 02 — cos θ2)

W = -pE(cos 180° – cos 0°)

=> W = -pE(-1 – (1)) = +2pE

Q.14. A 12 V battery is linked to three identical capacitors C1, C2, and C3, with capacitances of 6F each.

(i) What charge will be present on each capacitor?

(ii) What will the equivalent capacitance of the network be?

(iii) What will the energy stored in the network of capacitors be?

Ans: (i) C1 and C2 is series, make C4=3µF

Using the equation 1C4=1C1+1c2

(i) the potential that is available in C4 and C3

Charge in C3=Q3=C3V

=610-612=72C

Charge in C4=Q4=C4V

310-612=36C

∴Here the charge on CC1 and C2 will be the same, i.e., 36C

(ii) C4 and C4 are parallel to the source

∴Ceq=3+6=9F

(iii) Energy stored=12Ceq V²

=12(910-6)122

=64810-6=6.4810-4 joule.

### Class 12 Physics Chapter 2 Important Questions

Q.15. A battery supplies power to a parallel plate capacitor. After some time, the battery is removed, and a dielectric slab is put between the plates with a thickness equal to the plate separation. Find the following:

(i) the capacitance of the capacitor

(ii) the electric field in between the plates

(iii) how the energy stored in the capacitor will be affected

Ans. Take C as the capacitance and take V to be the potential difference.

Q = CV will be the charge on the capacitor plates

E = Vd will be the energy stored between the plates

En = Q22C or 12CV²

When the battery has disconnected, the dielectric (k) will be introduced

Then, the new volts of charge will be Q’= Q Capacitance C’= KC

Potential V’= QKC = VK

(i) New Capacitance is K times its original.

The new electric field E will be = V’d=VKd=EKi.e., 1K times the original field.

New energy = Q22C’ = Q22KC = 1K(En), i.e., 1K time the original energy

Q.16. A parallel plate capacitor is charged to a potential difference V. Each plate has an area of A and a distance of d. The battery that is used for charging is still attached. Between the plates is now a dielectric slab with the dimensions d, where k is the dielectric constant. What alterations, if any, will be made to

(i) the charge on plates?

(ii) the electric field intensity that would exist between the plates?

(iii) the capacitance of the capacitor?

Ans. The plate area of a parallel plate capacitor’s either plate equals A.

Potential difference between the plates equals V, and the distance between the plates equals d.

∴ Initially capacitance, C = 0Ad

Charge on the plate, Q – CV

When a dielectric slab of thickness denoted with “d” and dielectric constant denoted with “k” is placed between the plates, the potential difference between the plates remains unchanged (V’ = V) as long as the battery is connected throughout.

(i) New charge on the plates, Q’= C’V’= kCV = kQ

Hence, the charge will change to k times its original value.

(ii) Electric field intensity existing between the plates,

E’ = V’d=Vd = E

Hence, no change in the electric field intensity existing between the place of the capacitor.

(iii) The new capacitance of the capacitor will be

Q.17. Imagine there are two charges, 2 C and 2C, that are placed 6 cm apart from points A and B. Find the following.

(a) The equipotential surface of the system

(b) The electric field direction that exists at every point on the given surface

Ans. (a) This situation can be represented in the figure given below.

The plane where the total potential is zero everywhere is referred to as an equipotential surface. This plane is typical for line AB. Because the magnitude of the charges is the same throughout line AB, the plane is situated at its midpoint.

(b) Every point on this surface has an electric field that is normal to the plane and points in the direction of AB.

Q.18. Charge exists in a spherical conducting shell with an inner radius r1 and outer radius r2.

(a) In the middle of the shell is a charge q. What will the surface charge density of the shell’s interior and exterior surfaces be?

(b) Even if the shell is not spherical, but has any irregular shape, is the electric field within a cavity (with no charge) zero? Explain.

Ans. (a) A shell’s centre charge is designated as +q. So, the shell’s inner surface will generate a charge of size -q. As a result, the inner shell’s inner surface has a total charge of -q. The relation can be used to calculate the surface charge density at the shell’s inner surface.

1 = Total ChargeOuter Surface Area = -q4r12

2=Total ChargeOuter Surface Area = -q4r22

On the outside of the shell, a charge of +q is induced. The exterior of the shell is charged with a charge of Q magnitude. Therefore, the total charge on the shell’s exterior is Q+q.

(b) Yes, even though the electric field intensity existing inside a cavity is zetrot The shell has an uneven shape and is not spherical. Take a closed loop now, like so just a portion of it, along a field line, is within the hollow, and the remainder is inside the conductor. The field’s net effort in transporting a test charge over a closed loop. Due to the zero field inside the conductor, the loop will also be zero. Therefore, in whatever shape, the electric field is zero.

Q. A 50 V battery is coupled with a 12 pF capacitor. How much electrostatic energy does the capacitor have stored? Find the charge stored and the potential difference across each capacitor if a second 6 pF capacitor is placed in series with it and the same battery is connected across the combination.

Ans. The 12 pF capacitor’s capacity stores energy.

(i) U = 12CV²

= 12(1210-12)5050 J                (Given V = 50)

= 1.510-8 J

(ii) C = Equivalent capacitance of 12 pF and 6 pF in series can be represented by

1C=112+16=1+212

∴ C=4 pF

∴ The charge that will be stored across each capacitor will be

q=CV=(410-12)50 C=210-10 C

The charge that will be present on each capacitor 12 pF, and 6 pF is 210-10 C

∴ The potential difference existing across capacitor C1

V1=qC1=(210-12)1210-12 Volt=503V

The potential difference existing across capacitor C2

V2=qC2=(210-10)610-12 Volt=1003V

Q.19. There is a connection between two charged conducting spheres with radii a and b by a wire to another. How many electric fields are there on the surfaces of the

two spheres? Use the outcome to demonstrate why the charge density on the conductor’s sharp and pointed ends is higher than its flatter ones.

Portions?

Ans. Imagine “a” to be the sphere A’s radius, QA represents the sphere’s charge and CA represent the sphere’s capacitance. If “b” is a sphere B’s radius and QB is the sphere’s charge, let CB represent the sphere’s capacitance. The potential V of the two spheres will equalize since they are joined together by a wire.

Let EB and EA represent the electric fields of spheres B and A, respectively. Consequently, their ratio,

EAEB= QA40a2b240QB

EAEB=QAQBb2a2

But

QAQB=CAVCBV

CACB=ab

QAQB=ab

When we put the value obtained together, we get

EAEB = abb2a2 = ba

Hence, ba will be the electric field’s ratio at the surface.

Q.20. Given that there is a 0.5cm gap between the plates of a 2F parallel plate capacitor, what is the area of the plates? (You’ll see from your response why standard capacitors have a range of F or less. However, due to the extremely thin spacing between the electrolytic capacitors, electrolytic capacitors do have a significantly higher capacitance of 0.1F.)

Ans. The capacitance of a parallel capacitor can be written as V = 2F.

The gap between the two plates will be written as d = 0.5cm = 0.510-2m.

The Capacitance of a parallel plate capacitor can be represented by

0 = permittivity of free space = 8.8510-12C²N-1m²

A = 20.510-28.8510-12

A = 1130 km2

Therefore, the plate’s area is too big. The capacitance of the F range is taken to avoid this situation.

Q.21. (a) Give a definition of the dielectric constant with respect to the capacitance of a capacitor. What factor influences a parallel plate capacitor’s dielectric capacitance?

(b) To make the energy stored in the two cases equal, determine the ratio of the potential variations that must be applied across the

(i) parallel

(ii) series combination of two identical capacitors

Ans. Dielectric constant means the ratio of a capacitor’s capacitance in the case of the dielectric being filled between the plates with a capacitor’s capacitance in case of a vacuum present between the plates.

In K = CmCo = Capacitance of a capacitor when dielectric is in between the platesCapacitance of a capacitor with vacuum in between the plates

The following variables affect the capacitance of a parallel plate capacitor with a dielectric.

Cm= KA0d

(B)

Imagine the C to be the capacitance of each capacitor

CP = CCC+C = C2          ………….(in series)

Let the value of potential difference be  VP and Vs

Hence, UP = 12CPVP2 = 122CVP2= CVP2

US = 12CSVS2 = 12C2VS2 = CVS24

However, UP= US is already given,

CVP2 = CVS24

VP2VS2 = 14

VP: VS = 1:2

Q.22. Explain the construction, working, and the principle of a Van de Graaff generator using a labelled diagram. Also, mention its uses.

Ans. Van de Graaff generators are devices that can generate enormous potentials of the order of millions of volts.

Principle: In a hollow conductor, the charge is always located on the outer surface.

At the pointed ends of the conductors, the electric discharge in air or gas occurs easily.

How is it constructed?

It is made up of a big metallic hollow spherical S set on two insulating columns.

A and B are connected by a rubber belt that runs over two pulleys with the aid of an electric motor, P1 and P2, in an unending fashion. B1 and B2 are two pointed metallic brushes. High tension battery provides a positive voltage to the lower brush B1.

Spray brush, with the higher brush B2 attached to the S-sphere’s inner portion.

Working:

Brush B1 produces ions when its positive potential is increased because of the impact of sharp edges. The resulting positive ions are then sprayed on the belt as a result of electrostatic attraction between positive ions and brush B1’s positive charge.

It is then raised by the belt as it moves. B2’s pointy end barely touches

The belt gathers the favourable change and moves it to the belt’s outer surface S sphere. The potential of the shell increases to millions of volts as a result of this ongoing process.

Applications:

Protons, deuterons, and other particles, among others, are accelerated to very high speeds and energy.

### Chapter 12 – Electrostatic Potential and Capacitance

These Class 12 Physics Chapter 2 Important Questions include the Important Questions mentioned in NCERT books. After going through these important questions, you will know all about electric charges and fields and will gain confidence to solve CBSE past years’ question papers.

### Electrostatic Potential

A electrostatic potential is the work required to shift a unit of positive charge from one location to another in an electric field when that charge is up against an electrostatic force with zero acceleration.

The electrostatic potential equation is:

Work done (W)/Charge = Electrostatic potential (V) (q). The SI unit for electrostatic potential is the volt (V).

The electrostatic potential is equal to 1JC-1 when a charge of 1 coulomb is pushed by an electric field in opposition to an electrostatic force.

### Electrostatic Potential Difference

The electrostatic potential difference between points A and B is the amount of effort required to transport a unit positive charge from point A to point B with no acceleration when doing so in the presence of an electrostatic force and an electric field.

The electrostatic potential difference is calculated as follows.

VB-VA=WABq

The following is another electrostatic potential difference formula.

VB-VA= bA Edl

The line integral of the electric field from point A to point B can be used to determine the potential difference between those two points.

The electrostatic potential is calculated when a point charge q is present at any location P and a distance r as follows.

V = 140qr

The electric potential at that location will be positive if the point charge is positive. In contrast, the electric potential at a point will be negative if the point charge is negative.

In an electric field, a positive charge moves from a higher potential to a lower potential. In contrast, when placed in an electric field, negative potential moves from lower potential to greater potential.

### The Formula of Electrostatic Potential Due to an Electric Dipole

The formula of electrostatic potential caused to an electric dipole at a point S is given below.

V=140scosr2

### The Formula of Electrostatic Potential of a Thin Charged Spherical Shell

The electrostatic potential at a point P inside a thin, charged, spherical shell of radius R and charge q have the following formula.

V = 140qR

The electrostatic potential at a point S on the surface of a thin charged spherical shell with charge q and radius R has the following formula.

V = 140qR

The electrostatic potential at a point S outside of a thin, charged spherical shell carrying charge q and the distance r between that point and the shell’s centre is calculated as follows.

In the case where r is larger than R, V = 140qr

### Electrostatic Potential Energy

The work that a charge q performs when it is transported from one location to another is saved as electrical potential energy.

The following equation describes the electrostatic potential energy in a system with two charges, q1 and q2.

U = 140q1q2r

The students must understand these key ideas and formulas in order to successfully complete the electrostatic potential and capacitance Class 12 Physics Chapter 2 Important Questions.

Q1.

Four capacitors are connected as shown in the figure. What is the equivalent capacitances between A and B?

Opt.

According to the given figure, capacitance in parallel is 12 ?

Ans.

According to the given figure, capacitance in parallel is 12 ¼F and 12 ¼F.
Therefore, Cp = 12 + 12 = 24 ¼F

## 1. What variables affect a parallel plate capacitor's capacitance?

• Area of plates
• The difference between the plates’ spacing and
• The nature of dielectric that exists between them

10 V

## 3. What will the geometrical shape be if an equipotential surface is caused by a single isolated charge?

The geometrical shape created for an equipotential surface caused by a single isolated charge is concentric circles.

P = Xe0E

## 5. How much work is involved in moving a point charge across an r-radius circular arc where another point charge is situated?

The work done in this case will be zero as it is an equipotential surface.

## 6. Justify the following statement. “For any charge configuration, equipotential surface through a point is normal to the electric field”.

An equipotential surface requires no work to move a charge over it. Hence, every point on the surface will be normal to the electric field.