Electrostatic Potential Capacitance Class 12 Notes MCQ Solutions

Electrostatic Potential and Capacitance Class 12 Notes (Complete Guide)

Electrostatic Potential Capacitance Class 12 is one of the most important topics in Physics for CBSE board exams and competitive exams like NEET and JEE. This chapter helps you understand how electric charges store energy, how potential is defined, and how capacitors work in real-life applications.

In this complete guide, you will get clear and simple concepts, important formulas, MCQs, assertion reason questions, case study-based questions, and NCERT exercise solutions—all in one place. This makes it easier for students to revise quickly and practice different types of questions asked in exams.

If you want to score high marks, mastering Electrostatic Potential Capacitance Class 12 with proper understanding and practice is essential. This page is designed to give you everything you need for strong concept clarity and exam preparation.

Electrostatic Potential and Capacitance – Key Points

1. Electrostatic Force

Electrostatic force is a conservative force.
The work done by an external force (equal and opposite to electrostatic force) in moving a charge qqq from point RRR to point PPP is:W=q(VP−VR)W = q (V_P – V_R)

This work represents the change in potential energy of the charge.

2. Electric Potential

Electric potential at a point is defined as the work done per unit charge in bringing a test charge from infinity to that point.

  • Potential is defined up to an arbitrary constant
  • Only potential difference is physically meaningful

If potential at infinity is zero, then potential due to a point charge QQQ at distance rrr is:(Point charge potential)\text{(Point charge potential)}(Point charge potential)

V=14πε0QrV = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r}

3. Potential Due to Electric Dipole

The potential at a point due to a dipole of moment p⃗\vec{p}​ is:

V=14πε0p⃗⋅r^r2V = \frac{1}{4\pi\varepsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2}

This formula is valid when distance r≫ar \gg a

4. Superposition Principle

The total potential due to multiple charges is the sum of individual potentials:V=14πε0(q1r1P+q2r2P+⋯+qnrnP)V = \frac{1}{4\pi \varepsilon_0} \left( \frac{q_1}{r_{1P}} + \frac{q_2}{r_{2P}} + \cdots + \frac{q_n}{r_{nP}} \right)

5. Equipotential Surface

  • A surface where potential is constant
  • For a point charge → concentric spheres
  • Electric field:
    • Always perpendicular to equipotential surface
    • Directed towards maximum decrease in potential

6. Potential Energy of Charges

Potential energy of two charges q1q_1​ and q2q_2​ separated by distance r12r_{12}(Potential energy of two charges)\text{(Potential energy of two charges)}(Potential energy of two charges)

U=14πε0q1q2r12U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}}

7. Potential Energy in External Field

  • For a charge: U=qVU = qV
  • For a dipole in uniform electric field: U=−p⃗⋅E⃗U = -\vec{p} \cdot \vec{E}

8. Properties of Conductors

  • Electric field inside conductor = 0
  • Charges exist only on the surface
  • Electric field just outside: E=σε0E = \frac{\sigma}{\varepsilon_0}
  • Potential is constant inside and on the surface
  • Inside a cavity (no charge) → Electric field = 0

9. Capacitor

A capacitor consists of two conductors separated by an insulator.

  • Capacitance: C=QVC = \frac{Q}{V}
  • Unit: Farad (F)

For parallel plate capacitor:(Parallel plate capacitance)\text{(Parallel plate capacitance)}(Parallel plate capacitance)

C=ε0AdC = \varepsilon_0 \frac{A}{d}

10. Dielectric Effect

When dielectric is inserted:

  • Electric field reduces
  • Capacitance increases

C=KC0C = K C_0

Where KK = dielectric constant

11. Combination of Capacitors

Series Combination:1C=1C1+1C2+1C3+⋯\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \cdots

Parallel Combination:C=C1+C2+C3+⋯C = C_1 + C_2 + C_3 + \cdots

12. Energy Stored in Capacitor

(Energy stored in capacitor)\text{(Energy stored in capacitor)}(Energy stored in capacitor)

U=12CV2=12QV=Q22CU = \frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{Q^2}{2C}

  • Energy density:

u=12ε0E2u = \frac{1}{2} \varepsilon_0 E^2

MCQs on Electrostatic Potential and Capacitance (Class 12)

1. Electrostatic potential is a:

A. Vector quantity
B. Scalar quantity
C. Tensor quantity
D. None

Answer: B. Scalar quantity

2. Unit of electric potential is:

A. Joule
B. Newton
C. Volt
D. Coulomb

Answer: C. Volt

3. Potential at infinity is generally taken as:

A. 1
B. 10
C. 0
D. Infinite

Answer: C. 0

4. Equipotential surfaces are always:

A. Parallel to electric field
B. Perpendicular to electric field
C. Circular
D. Random

Answer: B. Perpendicular to electric field

5. Capacitance depends on:

A. Charge only
B. Voltage only
C. Geometry of conductors
D. Resistance

Answer: C. Geometry of conductors

6. In a conductor, electric field inside is:

A. Maximum
B. Minimum
C. Zero
D. Infinite

Answer: C. Zero

7. Capacitance of parallel plate capacitor increases when:

A. Distance increases
B. Area decreases
C. Dielectric is inserted
D. Voltage increases

Answer: C. Dielectric is inserted

8. Energy stored in capacitor depends on:

A. Charge
B. Voltage
C. Capacitance
D. All of these

Answer: D. All of these

9. In series combination, total capacitance is:

A. Greater
B. Smaller
C. Equal
D. Infinite

Answer: B. Smaller

10. Electric field lines are always:

A. Circular
B. Parallel
C. Perpendicular to equipotential surface
D. Random

Answer: C. Perpendicular to equipotential surface

Numerical Problems on Electrostatic Potential and Capacitance

1. Potential Due to Point Charge

Find the potential at a distance of 2 m from a charge of 4 μC.

V=14πε0QrV = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r}

Solution:

Given:
Q=4×10−6CQ = 4 \times 10^{-6} C

V=9×109×4×10−62V = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{2}V=18000 VV = 18000 \, V

2. Capacitance of Parallel Plate Capacitor

Find capacitance if area = 2 m² and distance = 0.01 m.

C=ε0AdC = \varepsilon_0 \frac{A}{d}

Solution:C=8.85×10−12×20.01C = 8.85 \times 10^{-12} \times \frac{2}{0.01}C=1.77×10−9 FC = 1.77 \times 10^{-9} \, F

3. Energy Stored in Capacitor

Find energy stored when C=2μFC = 2 \mu FV=10VV = 10 V

U=12CV2U = \frac{1}{2}CV^2

Solution:U=12×2×10−6×(10)2U = \frac{1}{2} \times 2 \times 10^{-6} \times (10)^2U=1×10−4 JU = 1 \times 10^{-4} \, J

4. Series Combination

Find equivalent capacitance of 2 μF and 3 μF in series.1C=12+13\frac{1}{C} = \frac{1}{2} + \frac{1}{3}1C=56\frac{1}{C} = \frac{5}{6}C=1.2 μFC = 1.2 \, \mu F

5. Parallel Combination

Find equivalent capacitance of 2 μF and 3 μF in parallel.C=2+3=5 μFC = 2 + 3 = 5 \, \mu F

Assertion–Reason Questions on Electrostatic Potential and Capacitance (Class 12)

Directions:

For each question, choose the correct option:
A. Both Assertion and Reason are true, and Reason is the correct explanation
B. Both Assertion and Reason are true, but Reason is not the correct explanation
C. Assertion is true, but Reason is false
D. Assertion is false, but Reason is true

1.

Assertion (A): Electrostatic force is a conservative force.
Reason (R): Work done depends only on initial and final positions.

Answer: A

2.

Assertion (A): Electric potential is a scalar quantity.
Reason (R): It has magnitude but no direction.

Answer: A

3.

Assertion (A): Potential at infinity is taken as zero.
Reason (R): Absolute value of potential is not important.

Answer: A

4.

Assertion (A): Equipotential surfaces are perpendicular to electric field lines.
Reason (R): Work done along equipotential surface is zero.

Answer: A

5.

Assertion (A): Electric field inside a conductor is zero.
Reason (R): Charges reside only on the surface.

Answer: A

6.

Assertion (A): Capacitance depends on charge stored.
Reason (R): Capacitance depends on geometry of conductors.

Answer: D

7.

Assertion (A): In series combination, capacitance decreases.
Reason (R): Effective distance between plates increases.

Answer: A

8.

Assertion (A): In parallel combination, capacitance increases.
Reason (R): Effective plate area increases.

Answer: A

9.

Assertion (A): Potential inside a conductor is constant.
Reason (R): Electric field inside is zero.

Answer: A

10.

Assertion (A): Energy stored in a capacitor is proportional to square of voltage.
Reason (R): Energy is given by:

U=12CV2U = \frac{1}{2}CV^2

Answer: A

11.

Assertion (A): Dielectric increases capacitance.
Reason (R): It reduces electric field inside capacitor.

Answer: A

12.

Assertion (A): Potential due to a point charge decreases with distance.
Reason (R): Potential is inversely proportional to distance.

Answer: A

13.

Assertion (A): Work done in moving a charge on an equipotential surface is zero.
Reason (R): Potential difference is zero.

Answer: A

14.

Assertion (A): Capacitance is independent of voltage.
Reason (R): It depends only on geometry.

Answer: A

15.

Assertion (A): Electric field is maximum where potential is constant.
Reason (R): Electric field is zero when potential is constant.

Answer: D

Case Study Questions on Electrostatic Potential and Capacitance (Class 12)

Case Study 1: Parallel Plate Capacitor

A student sets up a parallel plate capacitor with plate area AAA and separation ddd. The capacitor is connected to a battery and stores charge. Later, a dielectric slab is inserted between the plates.

The capacitance of a parallel plate capacitor is:

C=ε0AdC = \varepsilon_0 \frac{A}{d}

When a dielectric is inserted:C=KC0C = K C_0

Questions:

1. What happens to capacitance when dielectric is inserted?
A. Decreases
B. Increases
C. Remains same
D. Becomes zero

Answer: B

2. What is the effect on electric field inside the capacitor?
A. Increases
B. Decreases
C. No change
D. Becomes zero

Answer: B

3. If distance between plates is doubled, capacitance becomes:
A. Double
B. Half
C. Same
D. Zero

Answer: B

4. Capacitance depends on:
A. Charge
B. Voltage
C. Geometry
D. Time

Answer: C

5. If battery remains connected, charge on capacitor will:
A. Decrease
B. Increase
C. Remain same
D. Become zero

Answer: B

Case Study 2: Conductors and Equipotential Surfaces

A charged conductor is in electrostatic equilibrium. The electric field inside the conductor is zero and charges are present only on the surface.

Questions:

1. Why is electric field zero inside conductor?
A. No charges present
B. Charges cancel internal field
C. Field is infinite
D. Potential is zero

Answer: B

2. Potential inside a conductor is:
A. Zero
B. Maximum
C. Constant
D. Infinite

Answer: C

3. Equipotential surfaces are:
A. Parallel to electric field
B. Perpendicular to electric field
C. Random
D. Circular only

Answer: B

4. Work done along equipotential surface is:
A. Maximum
B. Minimum
C. Zero
D. Infinite

Answer: C

5. Direction of electric field is:
A. Along equipotential surface
B. Perpendicular to surface
C. Random
D. Circular

Answer: B

Case Study 3: Energy Stored in Capacitor

A capacitor is charged using a battery. Energy is stored in the electric field between the plates.

The energy stored is:

U=12CV2U = \frac{1}{2}CV^2

Questions:

1. If voltage is doubled, energy becomes:
A. Double
B. Half
C. Four times
D. Same

Answer: C

2. If capacitance increases, energy stored:
A. Decreases
B. Increases
C. Remains same
D. Zero

Answer: B

3. Energy is stored in:
A. Plates
B. Wires
C. Electric field
D. Battery

Answer: C

4. If charge is constant, increasing capacitance will:
A. Increase energy
B. Decrease energy
C. No change
D. Zero

Answer: B

5. Unit of energy stored is:
A. Volt
B. Coulomb
C. Joule
D. Farad

Answer: C

Case Study 4: Electric Potential and Point Charge

A point charge produces an electric field around it. The potential at a distance rrr is given by:

V=14πε0QrV = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r}

Questions:

1. If distance increases, potential:
A. Increases
B. Decreases
C. Constant
D. Zero

Answer: B

2. Potential depends on:
A. Distance
B. Charge
C. Medium
D. All

Answer: D

3. If charge is doubled, potential becomes:
A. Half
B. Same
C. Double
D. Zero

Answer: C

4. Potential is a:
A. Vector
B. Scalar
C. Tensor
D. Force

Answer: B

5. At infinity, potential is taken as:
A. 1
B. Infinite
C. Zero
D. Negative

Exercise 2 – Electrostatic Potential and Capacitance

2.1

Two charges 5×10−8 C5 \times 10^{-8} \, C and −3×10−8 C-3 \times 10^{-8} \, C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Answer

Let the zero potential point be at distance xxx from the positive charge.k⋅5x=k⋅316−x\frac{k \cdot 5}{x} = \frac{k \cdot 3}{16 – x}5(16−x)=3x5(16 – x) = 3x80−5x=3x⇒8x=8080 – 5x = 3x \Rightarrow 8x = 80x=10 cmx = 10 \, \text{cm}

2.2

A regular hexagon of side 10 cm has a charge 5 mC5 \, mC at each of its vertices. Calculate the electric potential at the centre of the hexagon.

Solution

In a regular hexagon, the distance from the centre to each vertex = side lengthr=10 cm=0.1 mr = 10 \, \text{cm} = 0.1 \, \text{m}

Potential due to one charge:V=14πε0⋅qrV = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r}

Total potential (since potential is scalar, we add directly):Vtotal=6×14πε0⋅qrV_{total} = 6 \times \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r}

Substitute values:V=6×9×109×5×10−60.1V = 6 \times \frac{9 \times 10^9 \times 5 \times 10^{-6}}{0.1}V=6×4.5×105V = 6 \times 4.5 \times 10^5V=2.7×106 VV = 2.7 \times 10^6 \, \text{V}

2.3

(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?

Two charges 2 μC2 \, \mu C and −2 μC-2 \, \mu C are placed at points A and B, 6 cm apart.

Ans: Answer

(a) Equipotential Surface

The perpendicular bisector of the line joining charges A and B is an equipotential surface.

This is because:

  • At every point on this line, distance from both charges is equal
  • Potentials due to +2 μC+2 \, \mu C and −2 μC-2 \, \mu C cancel each other
  • Hence, net potential = zero

(b) Direction of Electric Field

The electric field at every point on this surface is:

  • Perpendicular to the equipotential surface
  • Directed from positive charge to negative charge

So, the electric field is along the line from A to B, crossing the perpendicular bisector at right angles.

2.4

A spherical conductor of radius 12 cm has a charge of 1.6×10−7 C1.6 \times 10^{-7} \, C distributed uniformly on its surface. What is the electric field:

2.5

(a) inside the sphere?
(b) just outside the sphere?
(c) at a point 18 cm from the centre of the sphere?

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1 pF=10−12 F)(1 \, pF = 10^{-12} \, F)(1pF=10−12F). What will be the capacitance if:

  • the distance between the plates is reduced to half, and
  • the space between them is filled with a substance of dielectric constant 6?

2.6

Three capacitors, each of capacitance 9 pF, are connected in series.

(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

2.7

Three capacitors of capacitances 2 pF, 3 pF, and 4 pF are connected in parallel.

(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

2.8

In a parallel plate capacitor with air between the plates, each plate has an area of 6×10−3 m26 \times 10^{-3} \, m^2 and the distance between the plates is 3 mm.

Calculate:
(a) the capacitance of the capacitor
(b) the charge on each plate if the capacitor is connected to a 100 V supply

2.9

Explain what would happen if, in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (dielectric constant = 6) is inserted between the plates:

(a) while the voltage supply remains connected
(b) after the supply is disconnected

2.10

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

2.11

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

CLASS 12 PHYSICS CHAPTER 1 ELECTRIC CHARGES AND FIELD

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Answer: C

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