Chapter 2: Electrostatic Potential and Capacitance
Key Formulas:
• V = kQ/r
• U = kq₁q₂/r
• C = Q/V, C = ε₀A/d (parallel plate)
• Cseries = 1/C₁ + 1/C₂, Cparallel = C₁ + C₂
• Energy: U = Q²/2C = CV²/2 = QV/2
Q1. A charge of 2 μC is placed at the origin. Find the work done in bringing a charge of 4 μC from infinity to point (2m, 0, 0).
Solution
W = q₂V₁ = q₂(kq₁/r)
= (4 × 10−6)(9 × 109)(2 × 10−6)/2
= (4 × 10−6)(9 × 103)
= 36 × 10−3 J = 36 mJ
Q2. A parallel plate capacitor has capacitance 8 μF. What will be its capacitance if the separation between the plates is halved and a dielectric of constant 4 is introduced?
Solution
C₁ = ε₀A/d = 8 μF
C₂ = Kε₀A/(d/2) = 2K × (ε₀A/d) = 2K × C₁
= 2 × 4 × 8 = 64 μF
Q3. Two capacitors of 3 μF and 6 μF are connected in series. Find the equivalent capacitance and the charge on each if connected to a 9V battery.
Solution
1/Ceq = 1/3 + 1/6 = 3/6 = 1/2
Ceq = 2 μF
Q = CeqV = 2 × 9 = 18 μC
In series, charge on each capacitor = 18 μC
Q4. Find the potential at the centre of a uniformly charged spherical shell of radius R and total charge Q.
Solution
Potential is constant everywhere inside a charged shell.
Vcentre = Vsurface = kQ/R = Q/(4πε₀R)
Q5. A 10 μF capacitor is charged to 50V. The energy stored in the capacitor is:
Solution
U = ½CV² = ½(10 × 10−6)(50)²
= ½(10−5)(2500)
= 1.25 × 10−2 J = 12.5 mJ
Q6. Four identical capacitors are connected (i) all in series, (ii) all in parallel. Find the ratio of equivalent capacitances.
Solution
Let each capacitor = C
(i) Series: 1/Cs = 4/C ⇒ Cs = C/4
(ii) Parallel: Cp = 4C
Ratio Cs/Cp = (C/4)/(4C) = 1:16
Q7. Find the potential due to an electric dipole of dipole moment p at a point on its axis at distance r from the centre (r >> a).
Solution
For axial line (r >> a):
V = kp/r² = p/(4πε₀r²)
The potential falls off as 1/r² (faster than point charge's 1/r).
Q8. Two capacitors of 2 μF and 4 μF are connected in parallel across a 100V battery. After disconnecting the battery, the capacitors are reconnected in series. Find the potential difference across each.
Solution
In parallel: Q₁ = C₁V = 2 × 100 = 200 μC
Q₂ = C₂V = 4 × 100 = 400 μC
Total charge = 600 μC (same in series)
Ceq = (2 × 4)/(2 + 4) = 8/6 = 4/3 μF
Vtotal = Q/Ceq = 600/(4/3) = 450 V
V₁ = Q/C₁ = 600/2 = 300 V
V₂ = Q/C₂ = 600/4 = 150 V
Q9. A parallel plate capacitor is charged and then disconnected from the battery. What happens to (i) the charge (ii) the voltage (iii) the capacitance if the plate separation is doubled?
Solution
Since the battery is disconnected, charge Q remains constant.
(i) Charge: Remains unchanged (Q = constant)
(ii) Capacitance: C = ε₀A/d, so C₂ = C₁/2 (halved)
(iii) Voltage: V = Q/C, so V₂ = Q/(C₁/2) = 2V₁ (doubled)
Q10. An equipotential surface is always perpendicular to an electric field. Explain why.
Solution
The work done in moving a charge along an equipotential surface is zero.
dW = qE · dl = 0 for any displacement dl along the surface.
Since q ≠ 0 and dl is arbitrary along the surface, E must be perpendicular to the surface at every point.
Q11. A spherical conductor of radius 12 cm has a charge of 1.6 × 10−7 C on its surface. Find (i) the potential at a point 18 cm from the centre (ii) the potential at a point 8 cm from the centre.
Solution
(i) At r = 18 cm (outside): V = kQ/r
= (9 × 109)(1.6 × 10−7)/0.18
= 8000 V
(ii) At r = 8 cm (inside): V = Vsurface = kQ/R
= (9 × 109)(1.6 × 10−7)/0.12
= 12000 V (same as on surface)
Q12. Three capacitors of capacitances 3 μF, 4 μF, and 6 μF are connected in (a) series (b) parallel. Find the ratio of equivalent capacitances.
Solution
(a) Series: 1/Cs = 1/3 + 1/4 + 1/6 = (4+3+2)/12 = 9/12 = 3/4
Cs = 4/3 μF
(b) Parallel: Cp = 3 + 4 + 6 = 13 μF
Ratio Cs/Cp = (4/3)/13 = 4:39
Q13. The work done in moving a charge of 2 C across two points having a potential difference of 5 V is ______. Also find the energy stored in a capacitor of 5 μF charged to 10 μC.
Solution
W = qV = 2 × 5 = 10 J
Energy stored in capacitor:
U = Q²/2C = (10 × 10−6)²/2(5 × 10−6)
= 10−10/(10−5) = 10−5 J = 10 μJ
Q14. A dielectric of dielectric constant 4 fills the space between the plates of a parallel plate capacitor of capacitance 2 μF. Find (i) the free space capacitance (ii) the energy stored when charged to 200V.
Solution
(i) C = KC₀ ⇒ C₀ = C/K = 2/4 = 0.5 μF
(ii) U = ½CV² = ½(2 × 10−6)(200)²
= (10−6)(40000) = 0.04 J = 40 mJ