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

Important Questions with Answers for CBSE Class 12 Physics

1. What is electrostatic potential?

Answer: Electrostatic potential at a point is the work done in bringing a unit positive charge from infinity to that point, without acceleration, in an electric field.

2. Define the potential energy of a system of charges.

Answer: The potential energy of a system of charges is the work done in assembling the system of charges from infinity to their final positions in the field.

3. What is the relation between electric field and electrostatic potential?

Answer: The electric field (

$E$

E) is the negative gradient of the electrostatic potential (

$V$

V), i.e.,

$E = – \nabla V$

E=−∇V, or in terms of magnitude,

$E = – \frac{dV}{dr}$

E=−drdV​.

4. Derive an expression for the potential due to a point charge.

Answer: The electrostatic potential due to a point charge

$Q$

Q at a distance

$r$

r from the charge is given by:

$$V = \frac{kQ}{r}$$

V=rkQ​ where

$k$

k is Coulomb’s constant.

5. What is the concept of electric potential energy of a charge in an electric field?

Answer: The potential energy

$U$

U of a charge

$q$

q in an electric field is given by

$U = qV$

U=qV, where

$V$

V is the electric potential at the position of the charge.

6. Define capacitance.

Answer: Capacitance is the ability of a body to store charge per unit potential difference, and it is given by the formula

$C = \frac{Q}{V}$

C=VQ​, where

$Q$

Q is the charge and

$V$

V is the potential difference.

7. Derive the formula for the capacitance of a parallel plate capacitor.

Answer: The capacitance

$C$

C of a parallel plate capacitor is given by:

$$C = \epsilon_0 \frac{A}{d}$$

C=ϵ0​dA​ where

$A$

A is the area of one plate,

$d$

d is the separation between the plates, and

$\epsilon_0$

ϵ0​ is the permittivity of free space.

8. What is the energy stored in a capacitor?

Answer: The energy stored in a capacitor is given by:

$$U = \frac{1}{2} C V^2$$

U=21​CV2 or equivalently,

$U = \frac{Q^2}{2C}$

U=2CQ2​, where

$C$

C is the capacitance and

$V$

V is the potential difference.

9. How does a dielectric material affect the capacitance of a capacitor?

Answer: When a dielectric material is inserted between the plates of a capacitor, it increases the capacitance by a factor of the dielectric constant

$k$

k, so the new capacitance is

$C’ = kC$

C′=kC, where

$C$

C is the capacitance without the dielectric.

10. Explain the concept of dielectric breakdown.

Answer: Dielectric breakdown occurs when the electric field in the dielectric material exceeds a critical value, causing the dielectric to conduct and the capacitor to lose its ability to store charge.

11. What is the combination of capacitors in series? Derive the formula for total capacitance.

Answer: When capacitors are connected in series, the total capacitance

$C_t$

Ct​ is given by:

$$\frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_n}$$

Ct​1​=C1​1​+C2​1​+⋯+Cn​1​

12. What is the combination of capacitors in parallel? Derive the formula for total capacitance.

Answer: When capacitors are connected in parallel, the total capacitance

$C_t$

Ct​ is the sum of the individual capacitances:

$$C_t = C_1 + C_2 + \dots + C_n$$

Ct​=C1​+C2​+⋯+Cn​

13. What is the effect of a dielectric material on the electric field inside a capacitor?

Answer: The presence of a dielectric material decreases the electric field inside the capacitor because it polarizes in the presence of the electric field and partially cancels it out, reducing the effective field.

14. Derive the expression for the energy stored in a capacitor when it is charged by a battery.

Answer: The energy stored in a capacitor when charged by a battery is:

$$U = \frac{1}{2} C V^2$$

U=21​CV2 where

$C$

C is the capacitance and

$V$

V is the potential difference across the capacitor.

15. What is the relation between charge, capacitance, and potential difference?

Answer: The relationship between charge

$Q$

Q, capacitance

$C$

C, and potential difference

$V$

V is given by the equation:

$$Q = C \times V$$

Q=C×V

16. Question: How does the energy stored in a capacitor change when the distance between the plates is increased?

Answer:
When the distance between the plates of a capacitor is increased, the capacitance decreases. Since the energy stored in a capacitor is given by the formula

$U = \frac{1}{2} C V^2$

U=21​CV2, a decrease in capacitance leads to a decrease in the energy stored, provided the voltage remains constant.

17. Question: What is the effect of a dielectric material on the potential difference across the plates of a capacitor?

Answer:
When a dielectric material is inserted between the plates of a capacitor, the potential difference across the plates decreases for a given amount of charge. This is because the dielectric reduces the electric field between the plates, thus lowering the voltage needed to store the same amount of charge.

18. Question: What is the dielectric constant, and how does it affect the capacitance of a capacitor?

Answer:
The dielectric constant

$k$

k is a measure of a material’s ability to reduce the electric field between the plates of a capacitor. The capacitance of a capacitor increases by a factor of

$k$

k when a dielectric material is inserted between the plates. The new capacitance becomes

$C’ = kC$

C′=kC, where

$C$

C is the capacitance without the dielectric.

19. Question: Derive the expression for the electric field between the plates of a parallel plate capacitor.

Answer:
The electric field

$E$

E between the plates of a parallel plate capacitor is given by:

 

$$E = \frac{\sigma}{\epsilon_0}$$

E=ϵ0​σ​

where

$\sigma$

σ is the surface charge density (

$\sigma = \frac{Q}{A}$

σ=AQ​, where

$Q$

Q is the charge on each plate and

$A$

A is the area of the plate), and

$\epsilon_0$

ϵ0​ is the permittivity of free space.

For a parallel plate capacitor, the electric field is uniform between the plates.

20. Question: What is the relationship between the potential energy and the charge on a capacitor?

Answer:
The potential energy

$U$

U stored in a capacitor is related to the charge

$Q$

Q and the capacitance

$C$

C by the equation:

 

$$U = \frac{Q^2}{2C}$$

U=2CQ2​

This shows that for a given capacitance, the energy stored increases as the charge on the capacitor increases.

21. Question: How do you calculate the total energy stored in a system of capacitors connected in parallel?

Answer:
For capacitors connected in parallel, the total energy stored in the system is the sum of the energies stored in each individual capacitor. If

$C_1, C_2, \dots, C_n$

C1​,C2​,…,Cn​ are the capacitances of the capacitors connected in parallel, the total energy is given by:

 

$$U_{\text{total}} = \frac{1}{2} \sum_{i=1}^{n} C_i V^2$$

Utotal​=21​i=1∑n​Ci​V2

where

$V$

V is the potential difference across each capacitor (since they are in parallel, all capacitors have the same voltage).

22. Question: What is the energy density of an electric field?

Answer:
The energy density

$u$

u of an electric field is the energy stored per unit volume of the field and is given by the formula:

 

$$u = \frac{1}{2} \epsilon_0 E^2$$

u=21​ϵ0​E2

where

$E$

E is the electric field and

$\epsilon_0$

ϵ0​ is the permittivity of free space.

23. Question: How does the charge on a capacitor change when the dielectric constant is increased?

Answer:
When the dielectric constant of the material between the plates of a capacitor is increased, the capacitance increases, and thus for a given potential difference, the charge stored on the capacitor increases. This is because the relationship between charge, capacitance, and voltage is

$Q = C \times V$

Q=C×V, and an increase in capacitance leads to an increase in charge for the same voltage.

24. Question: What is the effect of increasing the voltage on the energy stored in a capacitor?

Answer:
The energy stored in a capacitor is proportional to the square of the voltage. As the voltage increases, the energy stored in the capacitor increases by a factor of

$V^2$

V2. The energy stored is given by:

 

$$U = \frac{1}{2} C V^2$$

U=21​CV2

So, doubling the voltage will increase the energy stored by a factor of four.

25. Question: What happens to the capacitance of a capacitor if it is disconnected from the battery and the dielectric material is inserted?

Answer:
If a capacitor is disconnected from the battery, the charge on the plates remains constant. When a dielectric material is inserted, the capacitance increases due to the dielectric constant

$k$

k, but the potential difference across the plates decreases. Since the charge remains the same, the energy stored in the capacitor also changes accordingly.

26. Question: How does the energy stored in a capacitor change if the capacitor is charged by a constant current?

Answer:
When a capacitor is charged by a constant current, the voltage across the capacitor increases linearly with time. The energy stored in the capacitor at any instant is given by:

 

$$U = \frac{1}{2} C V^2$$

U=21​CV2

Since the voltage increases with time, the energy stored also increases, and the rate of increase in energy depends on the current and the capacitance.

27. Question: What is the role of a capacitor in a circuit?

Answer:
A capacitor stores electrical energy and can release it when needed. It is used in circuits to smooth out fluctuations in voltage, filter signals, store energy temporarily, and in timing applications. It can also block direct current (DC) while allowing alternating current (AC) to pass through.

Very Short Answer Questions with Answers for CBSE Class 12 Physics

1. What is electrostatic potential?

Answer: Electrostatic potential is the work done in bringing a unit positive charge from infinity to a point in an electric field without acceleration.

2. What is capacitance?

Answer: Capacitance is the ability of a capacitor to store charge per unit potential difference, given by

$C = \frac{Q}{V}$

C=VQ​.

3. What is the formula for the potential due to a point charge?

Answer: The potential due to a point charge

$Q$

Q at a distance

$r$

r is

$V = \frac{kQ}{r}$

V=rkQ​, where

$k$

k is Coulomb’s constant.

4. What is the unit of capacitance?

Answer: The unit of capacitance is the farad (F), where

$1$

1 farad =

$1$

1 coulomb per volt.

5. What is the energy stored in a capacitor?

Answer: The energy stored in a capacitor is

$U = \frac{1}{2} C V^2$

U=21​CV2, where

$C$

C is capacitance and

$V$

V is the potential difference.

6. What is the relation between electric field and electrostatic potential?

Answer: The electric field

$E$

E is the negative gradient of the electrostatic potential:

$E = – \frac{dV}{dr}$

E=−drdV​.

7. What is the effect of a dielectric on capacitance?

Answer: A dielectric increases the capacitance of a capacitor by a factor of the dielectric constant

$k$

k, so

$C’ = kC$

C′=kC.

8. What happens to the capacitance of capacitors in series?

Answer: In series, the total capacitance

$C_t$

Ct​ is less than the capacitance of any individual capacitor:

$\frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2} + \dots$

Ct​1​=C1​1​+C2​1​+…

9. What happens to the capacitance of capacitors in parallel?

Answer: In parallel, the total capacitance

$C_t$

Ct​ is the sum of the individual capacitances:

$C_t = C_1 + C_2 + \dots$

Ct​=C1​+C2​+…

10. What is the dielectric constant?

Answer: The dielectric constant

$k$

k is a measure of a material’s ability to reduce the electric field between the plates of a capacitor.

11. What is the formula for the energy density of an electric field?

Answer: The energy density

$u$

u of an electric field is

$u = \frac{1}{2} \epsilon_0 E^2$

u=21​ϵ0​E2, where

$\epsilon_0$

ϵ0​ is the permittivity of free space and

$E$

E is the electric field.

12. What is the unit of electric potential?

Answer: The unit of electric potential is the volt (V), where

$1$

1 volt =

$1$

1 joule per coulomb.

13. What is the formula for total energy stored in a parallel combination of capacitors?

Answer: The total energy stored in a parallel combination of capacitors is the sum of the energy stored in each capacitor:

 

$$U_{\text{total}} = \sum \frac{1}{2} C_i V^2$$

Utotal​=∑21​Ci​V2

14. What is the effect of increasing the distance between plates in a parallel plate capacitor?

Answer: Increasing the distance between plates decreases the capacitance, as

$C = \frac{\epsilon_0 A}{d}$

C=dϵ0​A​, where

$d$

d is the distance.

15. What is the formula for the electric field between the plates of a parallel plate capacitor?

Answer: The electric field between the plates of a parallel plate capacitor is

$E = \frac{\sigma}{\epsilon_0}$

E=ϵ0​σ​, where

$\sigma$

σ is the surface charge density.