Exercises
1. Two charges 5 × 10⁻⁸ C and –3 × 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
Place q₁ = 5 × 10⁻⁸ C at the origin and q₂ = –3 × 10⁻⁸ C at x = 16 cm. Potential is zero where kq₁/r₁ + kq₂/r₂ = 0, i.e., 5/r₁ = 3/r₂.
Point between the charges (at distance x from q₁): 5/x = 3/(16 – x)
⇒ 80 – 5x = 3x
⇒ x = 10 cm.
Point outside, beyond the negative charge: 5/x = 3/(x – 16)
⇒ 5x – 80 = 3x
⇒ x = 40 cm.
So V = 0 at 10 cm and at 40 cm from the positive charge, both on the side of the negative charge. (No such point exists beyond the positive charge, because there the larger charge is always nearer.)
2. A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.
Answer
In a regular hexagon, the distance from the centre to each vertex equals the side, r = 0.1 m.
V = 6 × kq/r
⇒ V = 6 × (9 × 10⁹ × 5 × 10⁻⁶)/0.1
⇒ V = 6 × 4.5 × 10⁵
⇒ V = 2.7 × 10⁶ V
3. Two charges 2 μC and –2 μC are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?
Answer
(a) The plane passing through the midpoint of AB and perpendicular to AB. Every point on it is equidistant from the equal and opposite charges, so V = kq/r – kq/r = 0 everywhere on it.
(b) The field is always normal to an equipotential surface. So at every point of this plane the field is perpendicular to the plane, i.e., parallel to AB, pointing from A (positive charge) towards B (negative charge).
4. A spherical conductor of radius 12 cm has a charge of 1.6 × 10⁻⁷ C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?
Answer
(a) Inside a conductor the electrostatic field is zero: E = 0.
(b) Just outside, r = R = 0.12 m: E = kq/R²
⇒ (9 × 10⁹ × 1.6 × 10⁻⁷)/(0.12)²
⇒ 1440/0.0144
⇒ E = 1.0 × 10⁵ N/C, radially outward.
(c) r = 0.18 m: E = kq/r²
⇒ 1440/(0.18)²
⇒ 1440/0.0324
⇒ E ≈ 4.4 × 10⁴ N/C, radially outward.
5. A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1 pF = 10⁻¹² F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Answer
C = ε₀A/d = 8 pF. New capacitance C′ = Kε₀A/(d/2)
⇒ C′ = 2K × C
⇒ C′ = 2 × 6 × 8 pF
⇒ C′ = 96 pF
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?
Answer
(a) 1/C = 1/9 + 1/9 + 1/9 = 3/9
⇒ C = 3 pF
(b) In series each capacitor carries the same charge, and since they are identical the voltage divides equally: V each = 120/3 = 40 V.
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.
Answer
(a) C = 2 + 3 + 4 = 9 pF
(b) In parallel each capacitor has V = 100 V, Q = CV:
Q₁ = 2 × 10⁻¹² × 100 = 2 × 10⁻¹⁰ C; Q₂ = 3 × 10⁻¹² × 100 = 3 × 10⁻¹⁰ C; Q₃ = 4 × 10⁻¹² × 100 = 4 × 10⁻¹⁰ C
8. In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10⁻³ m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Answer
C = ε₀A/d
⇒ C = (8.854 × 10⁻¹² × 6 × 10⁻³)/(3 × 10⁻³)
⇒ C = 17.7 × 10⁻¹² F = 17.7 pF
Q = CV
⇒ Q = 17.7 × 10⁻¹² × 100
⇒ Q = 1.77 × 10⁻⁹ C (+ on one plate, – on the other)
9. Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected.
Answer
With the mica filling the gap, capacitance becomes C′ = KC = 6 × 17.7 pF = 106.2 pF.
(a) Supply connected: voltage stays at 100 V. Charge increases: Q′ = C′V
⇒ 106.2 × 10⁻¹² × 100
⇒ Q′ ≈ 1.06 × 10⁻⁸ C (six times the earlier charge; the extra charge flows in from the supply).
(b) Supply disconnected: charge stays at 1.77 × 10⁻⁹ C. Voltage drops: V′ = Q/C′ = V/K
⇒ 100/6
⇒ V′ ≈ 16.6 V.
10. A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?
Answer
U = ½CV²
⇒ U = ½ × 12 × 10⁻¹² × (50)²
⇒ U = 6 × 10⁻¹² × 2500
⇒ U = 1.5 × 10⁻⁸ J
11. A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Answer
Initial energy: U₁ = ½CV²
⇒ ½ × 600 × 10⁻¹² × (200)²
⇒ U₁ = 1.2 × 10⁻⁵ J
Charge Q = CV = 600 × 10⁻¹² × 200 = 1.2 × 10⁻⁷ C, shared by two equal capacitors in parallel (total 1200 pF).
Common voltage V′ = Q/(2C)
⇒ 1.2 × 10⁻⁷/(1200 × 10⁻¹²)
⇒ V′ = 100 V
Final energy: U₂ = ½ × 1200 × 10⁻¹² × (100)²
⇒ U₂ = 6 × 10⁻⁶ J
Energy lost = U₁ – U₂ = 1.2 × 10⁻⁵ – 6 × 10⁻⁶ = 6 × 10⁻⁶ J (dissipated as heat and electromagnetic radiation in the connecting wires).
Important Extra Questions
Define electrostatic potential at a point. Is it a scalar or a vector?
Answer
It is the work done by an external force in bringing a unit positive test charge from infinity to that point, slowly (without acceleration). It is a scalar; SI unit volt (J/C).
Why is no work done in moving a charge on an equipotential surface?
Answer
W = q(V_B – V_A). On an equipotential surface V_B = V_A, so W = 0. Also, the field is perpendicular to the surface, so the force has no component along the displacement.
Write the relation between electric field and potential.
Answer
E = –dV/dr: the field points in the direction of steepest decrease of potential, and its magnitude equals the change in potential per unit displacement normal to the equipotential surface.
What is the potential energy of an electric dipole in a uniform field?
Answer
U(θ) = –pE cos θ = –p·E. It is minimum (–pE) when the dipole is parallel to the field (stable equilibrium) and maximum (+pE) when antiparallel.
Why is the electric field zero inside a charged conductor, and where does the excess charge reside?
Answer
Free electrons rearrange until the field inside vanishes; otherwise they would keep moving. By Gauss's law, any surface inside then encloses no net charge, so the excess charge resides only on the outer surface, and the potential is the same throughout the conductor.