Electrostatic Potential and Capacitance — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Electrostatic Potential and Capacitance" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Electrostatic Potential and Capacitance" — 8 important questions with detailed answers…
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NCERT Solutions for Class 12 Physics — Electrostatic Potential and Capacitance. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Electrostatic Potential and Capacitance — All 8 Questions Solved
Q1. Define electric potential. Write expression for potential due to point charge.
Electric Potential (V): Work done per unit positive charge in bringing it from infinity to that point in electric field.
V = W/q₀ (potential is scalar quantity)
SI Unit: Volt (V) or J/C
Potential due to point charge q at distance r:
V = kq/r = q/(4πε₀r)
where k = 9 × 10⁹ N⋅m²/C²
ε₀ = 8.85 × 10⁻¹² F/m
Note: V = 0 at infinity (reference point)
Q2. What is potential difference? A charge of 2 C is moved from point A to point B, and 20 J of work is done. Find potential difference between A and B.
Potential Difference (V_AB): Difference in electric potential between two points.
V_AB = V_A - V_B = W_AB/q
where W_AB = work done by external agent to move charge q from B to A
Given: q = 2 C
W = 20 J
Step 1: V_AB = W/q = 20/2 = 10 V
Final Answer: Potential difference = 10 V
This means point A is at 10 V higher potential than point B. Work done against electric field = 20 J.
Q3. Define capacitance. Write expression for capacitance of parallel plate capacitor.
Capacitance (C): Ability of conductor to store charge for given potential difference.
C = Q/V (charge per unit potential difference)
SI Unit: Farad (F) = C/V
For Parallel Plate Capacitor:
C = ε₀εᵣA/d = ε₀A/d (in vacuum/air)
where ε₀ = 8.85 × 10⁻¹² F/m (permittivity of free space)
εᵣ = relative permittivity of dielectric
A = area of each plate
d = separation between plates
Note: Capacitance is independent of charge and voltage, depends only on geometry and material.
Q4. Two capacitors of 2 μF and 4 μF are connected in series across 600 V supply. Find charge on each capacitor and voltage across each.
Given: C₁ = 2 μF, C₂ = 4 μF, V = 600 V (series connection)
In series connection: Same charge on both capacitors
Step 1: Find equivalent capacitance
1/C_eq = 1/C₁ + 1/C₂ = 1/2 + 1/4 = 3/4
C_eq = 4/3 μF
Step 2: Total charge Q = C_eq × V = (4/3) × 600 = 800 μC
Step 3: Charge on each capacitor = 800 μC
(Q₁ = Q₂ = 800 μC since in series)
Step 4: Voltage across C₁
V₁ = Q/C₁ = 800/2 = 400 V
Step 5: Voltage across C₂
V₂ = Q/C₂ = 800/4 = 200 V
Verification: V₁ + V₂ = 400 + 200 = 600 V ✓
Final Answer: Charge on each = 800 μC; V₁ = 400 V; V₂ = 200 V
Q5. Derive expression for energy stored in a capacitor.
Consider charging capacitor from uncharged state to final charge Q.
At any instant, if charge is q, potential difference = q/C
Work done to add dq: dW = V⋅dq = (q/C)dq
Total work done:
W = ∫₀^Q (q/C)dq = (1/C)∫₀^Q q⋅dq = (1/C)[q²/2]₀^Q = Q²/(2C)
This work is stored as electrostatic potential energy.
Energy stored: U = Q²/(2C)
Alternative forms:
U = (1/2)QV (since Q = CV)
U = (1/2)CV² (substituting Q = CV)
where Q = charge on capacitor
V = final potential difference
C = capacitance
Energy density (energy per unit volume):
u = (1/2)ε₀E² (in terms of electric field E)
Q6. A parallel plate capacitor with plate area 100 cm² and separation 2 mm is charged to 100 V. Calculate: (a) Capacitance (b) Charge stored (c) Energy stored
Given: A = 100 cm² = 100 × 10⁻⁴ m² = 0.01 m²
d = 2 mm = 2 × 10⁻³ m
V = 100 V
ε₀ = 8.85 × 10⁻¹² F/m
(a) Capacitance:
C = ε₀A/d = (8.85 × 10⁻¹²) × (0.01) / (2 × 10⁻³)
C = 8.85 × 10⁻¹⁴ / (2 × 10⁻³)
C = 4.425 × 10⁻¹¹ F
C ≈ 44.25 pF
(b) Charge stored:
Q = CV = 4.425 × 10⁻¹¹ × 100
Q = 4.425 × 10⁻⁹ C
Q ≈ 4.425 nC
(c) Energy stored:
U = (1/2)CV² = (1/2) × 4.425 × 10⁻¹¹ × (100)²
U = 0.5 × 4.425 × 10⁻¹¹ × 10⁴
U = 2.2125 × 10⁻⁷ J
U ≈ 221.25 nJ
Final Answer: (a) C ≈ 44.25 pF (b) Q ≈ 4.425 nC (c) U ≈ 221.25 nJ
Q7. What is dielectric? How does dielectric affect capacitance? Write relation between ε₀, εᵣ, and ε.
Dielectric: Insulating material containing bound charges (atoms with electrons bound to nuclei).
Effect on Capacitance: When dielectric is inserted between plates, capacitance increases by factor εᵣ (relative permittivity).
C = εᵣC₀ = εᵣε₀A/d
where C₀ = ε₀A/d (capacitance in vacuum/air)
Relation between constants:
ε = εᵣε₀
where ε = absolute permittivity of material
εᵣ = relative permittivity (dielectric constant) ≥ 1
ε₀ = permittivity of free space = 8.85 × 10⁻¹² F/m
Physical reason: Electric field of dielectric opposes applied field, resulting in net reduced field inside. This allows more charge storage.
Typical εᵣ values:
Vacuum: εᵣ = 1
Air: εᵣ ≈ 1.0006
Paper: εᵣ ≈ 3-4
Mica: εᵣ ≈ 5-7
Water: εᵣ ≈ 80
Q8. Derive electric potential due to dipole on its equatorial line.
Consider dipole with charges +q and -q separated by 2a. Point P on equatorial line at distance r from center (r >> a).
Distance from +q to P: r₊ = √(r² + a²)
Distance from -q to P: r₋ = √(r² + a²)
Potential at P:
V = V₊ + V₋ = kq/r₊ - kq/r₋
Since r >> a, both distances approximately equal to r, difference is very small.
Using exact distances:
V = kq/√(r² + a²) - kq/√(r² + a²) = 0
Actually, on the equatorial line, the two charges are equidistant, so:
V = kq/√(r² + a²) - kq/√(r² + a²) = 0
Final Answer: V = 0 (on equatorial line of dipole)
Note: This is independent of distance. All points on equatorial line (perpendicular bisector) have zero potential because they are equidistant from +q and -q.
Electrostatic Potential and Capacitance — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the electrostatic potential at the center of a hollow conducting sphere carrying a uniform surface charge?
- A. Zero
- B. Same as the potential on the surface
- C. Inversely proportional to the radius
- D. Directly proportional to the charge density
Answer: B. Same as the potential on the surface — Inside a hollow conducting sphere, the electric field is zero. Therefore, the potential inside is constant and equal to the potential on its surface.
MCQ 2. Two point charges, +q and -q, are placed at (0, 0, -a) and (0, 0, a) respectively. What is the electrostatic potential energy of this system?
- A. -kq²/a
- B. -kq²/(2a)
- C. kq²/a
- D. kq²/(2a)
Answer: B. -kq²/(2a) — The potential energy of a system of two charges is given by U = k * q1 * q2 / r. Here, r = 2a, so U = k * (+q) * (-q) / (2a) = -kq²/(2a).
MCQ 3. Which of the following statements about equipotential surfaces is incorrect?
- A. Electric field lines are always perpendicular to equipotential surfaces.
- B. No work is done in moving a charge along an equipotential surface.
- C. Two equipotential surfaces can intersect each other.
- D. The potential is constant over an equipotential surface.
Answer: C. Two equipotential surfaces can intersect each other. — If two equipotential surfaces intersected, there would be two different potentials at the point of intersection, which is physically impossible.
MCQ 4. If the electric potential V in a region is given by V = 2x²y, what is the magnitude of the electric field at point (1, 1, 0)?
- A. 2√5 V/m
- B. 2√3 V/m
- C. 4 V/m
- D. 6 V/m
Answer: A. 2√5 V/m — The electric field E = -∇V. Calculating partial derivatives: ∂V/∂x = 4xy, ∂V/∂y = 2x², ∂V/∂z = 0. So, E = -(4xy î + 2x² ĵ). At (1,1,0), E = -(4î + 2ĵ). Magnitude |E| = √((-4)² + (-2)²) = √20 = 2√5 V/m.
MCQ 5. The capacitance of a capacitor depends on:
- A. The charge stored on its plates.
- B. The potential difference between its plates.
- C. The geometry of the capacitor and the dielectric medium between plates.
- D. The material of the plates.
Answer: C. The geometry of the capacitor and the dielectric medium between plates. — Capacitance is a geometric property of a capacitor, determined by the size, shape, separation of its conductors, and the dielectric material between them. It does not depend on Q or V.
MCQ 6. A parallel plate capacitor has plate area A and separation d. If the space between the plates is filled with a dielectric of relative permittivity ε_r, its capacitance is:
- A. ε_0 A / d
- B. ε_r ε_0 A / d
- C. ε_0 A / (ε_r d)
- D. d / (ε_0 A)
Answer: B. ε_r ε_0 A / d — The capacitance of a parallel plate capacitor in vacuum is C₀ = ε₀ A / d. When a dielectric of relative permittivity ε_r fills the space, the capacitance becomes C = ε_r C₀ = ε_r ε₀ A / d.
MCQ 7. Three capacitors, each of capacitance C, are connected in series. This combination is then connected in parallel with another capacitor of capacitance C. The equivalent capacitance of the entire arrangement is:
- A. C/4
- B. 4C/3
- C. 3C/4
- D. 2C
Answer: B. 4C/3 — For capacitors in series, 1/C_s = 1/C + 1/C + 1/C = 3/C, so C_s = C/3. This equivalent capacitance C_s is then in parallel with another capacitor C, so C_eq = C_s + C = C/3 + C = 4C/3.
MCQ 8. A capacitor of capacitance 20 μF is charged to a potential difference of 200 V. The energy stored in the capacitor is:
- A. 0.4 J
- B. 0.8 J
- C. 400 J
- D. 800 J
Answer: A. 0.4 J — Energy stored U = (1/2)CV². U = (1/2) * (20 * 10⁻⁶ F) * (200 V)² = (1/2) * 20 * 10⁻⁶ * 40000 = 0.4 J.
MCQ 9. A parallel plate capacitor is charged by a battery and then the battery is disconnected. A dielectric slab is then inserted between the plates. Which of the following statements is INCORRECT?
- A. The charge on the plates remains constant.
- B. The capacitance of the capacitor increases.
- C. The potential difference across the plates increases.
- D. The electric field between the plates decreases.
Answer: C. The potential difference across the plates increases. — When the battery is disconnected, the charge Q on the plates remains constant. Inserting a dielectric increases the capacitance (C=kC₀). Since Q=CV, and Q is constant while C increases, the potential difference V=Q/C must decrease.
MCQ 10. The electrostatic potential at a point due to a point charge 'q' is inversely proportional to:
- A. The distance from the charge (r)
- B. The square of the distance from the charge (r²)
- C. The cube of the distance from the charge (r³)
- D. The square root of the distance from the charge (√r)
Answer: A. The distance from the charge (r) — The electrostatic potential due to a point charge q at a distance r is given by V = kq/r, which shows it is inversely proportional to r.
MCQ 11. At a point on the equatorial plane of an electric dipole, the electric potential is:
- A. Zero
- B. Non-zero and positive
- C. Non-zero and negative
- D. Dependent on the distance from the center and the dipole moment
Answer: A. Zero — The electric potential due to a dipole at any point on its equatorial plane is zero because the potential due to the positive charge and the negative charge cancel each other out due to equal distance and opposite signs.
MCQ 12. What is the work done in moving a charge of 5 μC from a point A (potential 10 V) to a point B (potential -5 V)?
- A. 75 μJ
- B. -75 μJ
- C. 25 μJ
- D. -25 μJ
Answer: B. -75 μJ — Work done W = q * (V_B - V_A). W = (5 * 10⁻⁶ C) * (-5 V - 10 V) = (5 * 10⁻⁶) * (-15) = -75 * 10⁻⁶ J = -75 μJ.
MCQ 13. A spherical conductor of radius R is given a charge Q. What is its capacitance?
- A. 4πε₀R
- B. 4πε₀R²
- C. Q / (4πε₀R)
- D. Q / R
Answer: A. 4πε₀R — The potential on the surface of a spherical conductor of radius R with charge Q is V = Q / (4πε₀R). By definition, C = Q/V. Substituting V, we get C = Q / (Q / (4πε₀R)) = 4πε₀R.
MCQ 14. A parallel plate capacitor has area A and plate separation d. It is filled with two dielectric slabs, each of thickness d/2, with dielectric constants k₁ and k₂ respectively. The equivalent capacitance is:
- A. (k₁ + k₂) ε₀ A / (2d)
- B. 2ε₀ A / (d(k₁ + k₂))
- C. (2k₁k₂ / (k₁ + k₂)) ε₀ A / d
- D. ((k₁ + k₂) / (2k₁k₂)) ε₀ A / d
Answer: C. (2k₁k₂ / (k₁ + k₂)) ε₀ A / d — The two dielectric slabs act as two capacitors in series. C₁ = k₁ε₀A / (d/2) = 2k₁ε₀A / d and C₂ = k₂ε₀A / (d/2) = 2k₂ε₀A / d. For series combination, 1/C_eq = 1/C₁ + 1/C₂ = d / (2k₁ε₀A) + d / (2k₂ε₀A) = (d / (2ε₀A)) * (1/k₁ + 1/k₂) = (d / (2ε₀A)) * ((k₁ + k₂) / (k₁k₂)). Thus, C_eq = (2k₁k₂ / (k₁ + k₂)) ε₀ A / d.
MCQ 15. Two point charges of +10 μC and -10 μC are placed 10 cm apart. What is the potential at a point exactly midway between them?
- A. 0 V
- B. 1.8 * 10⁶ V
- C. -1.8 * 10⁶ V
- D. 9 * 10⁵ V
Answer: A. 0 V — The midpoint is equidistant (5 cm) from both charges. The potential due to the positive charge (kq/r) and the negative charge (-kq/r) will be equal in magnitude and opposite in sign, thus canceling each other out to give a net potential of 0 V.
MCQ 16. The SI unit of electrostatic potential is:
- A. Joule (J)
- B. Coulomb (C)
- C. Volt (V)
- D. Farad (F)
Answer: C. Volt (V) — The SI unit of electrostatic potential is Volt (V), which is defined as Joules per Coulomb (J/C).
MCQ 17. The plates of a parallel plate capacitor are charged to a potential difference V. If the distance between the plates is doubled while the charge on the plates remains constant, the potential difference between the plates will:
- A. Halve
- B. Double
- C. Remain the same
- D. Quadruple
Answer: B. Double — Capacitance C = ε₀A/d. If d is doubled, C becomes C/2. Since the charge Q remains constant (Q=CV), the new potential V' = Q/C' = Q/(C/2) = 2(Q/C) = 2V. So, the potential difference doubles.
MCQ 18. Two capacitors of capacitances C₁ and C₂ are charged to potentials V₁ and V₂ respectively. If they are connected in parallel, the common potential is:
- A. (C₁V₁ + C₂V₂) / (C₁ + C₂)
- B. (C₁V₁ - C₂V₂) / (C₁ + C₂)
- C. (C₁V₁ + C₂V₂) / (C₁ - C₂)
- D. (C₁V₁ - C₂V₂) / (C₁ - C₂)
Answer: A. (C₁V₁ + C₂V₂) / (C₁ + C₂) — When capacitors are connected in parallel, the total charge is conserved. Total initial charge Q_initial = C₁V₁ + C₂V₂. The total final charge Q_final = (C₁ + C₂)V_common. Equating them gives V_common = (C₁V₁ + C₂V₂) / (C₁ + C₂).
MCQ 19. The energy density of an electric field E in a vacuum is given by:
- A. (1/2)ε₀E²
- B. (1/2)E²
- C. ε₀E²
- D. 2ε₀E²
Answer: A. (1/2)ε₀E² — The energy density (energy per unit volume) of an electric field E in a vacuum is given by the formula u = (1/2)ε₀E².
MCQ 20. An electric dipole of dipole moment 'p' is placed in a uniform electric field 'E'. The potential energy of the dipole is minimum when the angle between 'p' and 'E' is:
- A. 0°
- B. 90°
- C. 180°
- D. 45°
Answer: A. 0° — The potential energy of an electric dipole in a uniform electric field is U = -p⋅E = -pE cosθ. For U to be minimum (most negative), cosθ must be maximum, which occurs when θ = 0°.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.