Electrostatic Potential and Capacitance

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Exercise 2.1Electrostatic Potential and Capacitors

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.

Solution

Let q₁ = 5 × 10⁻⁸ C at A and q₂ = –3 × 10⁻⁸ C at B, with AB = 16 cm = 0.16 m. Let P be a point on line AB where V = 0. Case 1: P between A and B (at distance x from A, so 0.16 – x from B): V = kq₁/x + kq₂/(0.16 – x) = 0 5 × 10⁻⁸/x = 3 × 10⁻⁸/(0.16 – x) 5(0.16 – x) = 3x 0.8 – 5x = 3x 0.8 = 8x x = 0.1 m = 10 cm from A Case 2: P outside AB, on the side of A (at distance x from A, beyond A): V = kq₁/x + kq₂/(0.16 + x) ... wait, outside on far side of A: Actually for P outside on far side of B (at distance x from B): V = kq₁/(0.16 + x) + kq₂/x = 0 5/(0.16 + x) = 3/x 5x = 3(0.16 + x) 5x = 0.48 + 3x 2x = 0.48 x = 0.24 m = 24 cm from B (on the far side of B from A) So V = 0 at: • 10 cm from the +5 × 10⁻⁸ C charge (between the charges) • 40 cm from the +5 × 10⁻⁸ C charge (24 cm beyond the –3 × 10⁻⁸ C charge)

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.

Solution

A regular hexagon has all vertices at equal distance from the centre. Side of hexagon = distance from centre to vertex = 10 cm = 0.1 m Potential at centre due to one charge: V₁ = kq/r = (9 × 10⁹ × 5 × 10⁻⁶) / 0.1 = 4.5 × 10⁵ V Total potential at centre (6 charges, all at same distance, all same charge): V = 6V₁ = 6 × 4.5 × 10⁵ = 2.7 × 10⁶ V The potential at the centre is 2.7 × 10⁶ V.

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?

Solution

(a) For two equal and opposite charges (+2μC and –2μC), the plane perpendicular to the line joining the charges and passing through the midpoint is the equipotential surface where V = 0. This is because at any point P on this plane: • Distance from +q equals distance from –q • V = kq/r + k(–q)/r = 0 The equipotential surface is the plane passing through the midpoint of AB, perpendicular to AB. (b) The electric field at every point on this plane is along the direction from the positive charge to the negative charge (i.e., along the line AB). Note: Electric field is always perpendicular to equipotential surfaces. Since the plane is the equipotential, E must be perpendicular to it — i.e., E is directed from A (+q) to B (–q).

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?

Solution

q = 1.6 × 10⁻⁷ C, R = 12 cm = 0.12 m (a) Inside the sphere (r < R): For a conducting sphere, E = 0 inside. (b) Just outside the sphere (r = R): E = kq/R² = σ/ε₀ = (9 × 10⁹ × 1.6 × 10⁻⁷) / (0.12)² = 1440 / 0.0144 = 10⁵ N/C Direction: radially outward (c) At r = 18 cm = 0.18 m (outside the sphere): The sphere acts as a point charge at the centre: E = kq/r² = (9 × 10⁹ × 1.6 × 10⁻⁷) / (0.18)² = 1440 / 0.0324 ≈ 4.44 × 10⁴ N/C Direction: radially outward

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 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?

Solution

Initial capacitance: C₁ = 8 pF = ε₀A/d New conditions: • Distance = d/2 (reduced by half) • Dielectric constant K = 6 New capacitance: C₂ = Kε₀A/(d/2) = 2Kε₀A/d = 2K × C₁ = 2 × 6 × 8 = 96 pF The new capacitance is 96 pF.

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?

Solution

(a) For capacitors in series: 1/C = 1/C₁ + 1/C₂ + 1/C₃ = 1/9 + 1/9 + 1/9 = 3/9 = 1/3 C = 3 pF (b) Charge on each capacitor (same in series): Q = CV = 3 × 10⁻¹² × 120 = 3.6 × 10⁻¹⁰ C Potential difference across each capacitor: V₁ = Q/C₁ = 3.6 × 10⁻¹⁰ / 9 × 10⁻¹² = 40 V Each capacitor has V = 40 V across it. (Check: 3 × 40 = 120 V ✓)

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.

Solution

(a) For capacitors in parallel: C = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9 pF (b) In parallel, voltage across each = 100 V Q₁ = C₁V = 2 × 10⁻¹² × 100 = 2 × 10⁻¹⁰ C = 200 pC Q₂ = C₂V = 3 × 10⁻¹² × 100 = 3 × 10⁻¹⁰ C = 300 pC Q₃ = C₃V = 4 × 10⁻¹² × 100 = 4 × 10⁻¹⁰ C = 400 pC

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?

Solution

A = 6 × 10⁻³ m², d = 3 mm = 3 × 10⁻³ m, ε₀ = 8.85 × 10⁻¹² F/m Capacitance: C = ε₀A/d = 8.85 × 10⁻¹² × 6 × 10⁻³ / 3 × 10⁻³ = 8.85 × 10⁻¹² × 2 = 17.7 × 10⁻¹² F = 17.7 pF Charge at V = 100 V: Q = CV = 17.7 × 10⁻¹² × 100 = 1.77 × 10⁻⁹ C ≈ 1.77 nC

Explain what would happen if in the capacitor given in Q8, 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.

Solution

Original: C₀ = 17.7 pF, V₀ = 100 V, Q₀ = 1.77 nC With mica (K = 6) inserted: New capacitance C = KC₀ = 6 × 17.7 = 106.2 pF (a) Supply remains connected (V = 100 V constant): • Voltage remains 100 V • New charge Q = CV = 106.2 × 10⁻¹² × 100 = 1.062 × 10⁻⁸ C ≈ 10.62 nC • Charge increases (additional charge flows from supply) • Electric field E = V/d = 100/(3 × 10⁻³) = 33,333 V/m (unchanged) • Energy stored = ½CV² = ½ × 106.2 × 10⁻¹² × 100² = 5.31 × 10⁻⁷ J (increases) (b) Supply disconnected (Q = Q₀ = 1.77 nC constant): • Charge remains 1.77 nC • New voltage V = Q/C = 1.77 × 10⁻⁹ / 106.2 × 10⁻¹² = 16.67 V (decreases) • Electric field E = V/d decreases • Energy stored = Q²/2C decreases (energy is given to the dielectric)

Q10

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

Solution

C = 12 pF = 12 × 10⁻¹² F, V = 50 V Energy stored: U = ½CV² = ½ × 12 × 10⁻¹² × (50)² = ½ × 12 × 10⁻¹² × 2500 = 6 × 10⁻¹² × 2500 = 1.5 × 10⁻⁸ J = 15 × 10⁻⁹ J = 15 nJ

Q11

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 this process?

Solution

Initial charge: Q = CV = 600 × 10⁻¹² × 200 = 1.2 × 10⁻⁷ C Initial energy: U₁ = ½CV² = ½ × 600 × 10⁻¹² × (200)² = ½ × 600 × 10⁻¹² × 40000 = 1.2 × 10⁻⁵ J After connection to uncharged 600 pF capacitor: Total capacitance = 600 + 600 = 1200 pF (parallel) Charge is conserved: Q_total = 1.2 × 10⁻⁷ C New voltage: V' = Q_total / C_total = 1.2 × 10⁻⁷ / 1200 × 10⁻¹² = 100 V Final energy: U₂ = ½C_total × V'² = ½ × 1200 × 10⁻¹² × (100)² = ½ × 1200 × 10⁻¹² × 10000 = 6 × 10⁻⁶ J Energy lost = U₁ – U₂ = 1.2 × 10⁻⁵ – 6 × 10⁻⁶ = 12 × 10⁻⁶ – 6 × 10⁻⁶ = 6 × 10⁻⁶ J Energy lost = 6 × 10⁻⁶ J = 6 μJ (This energy is dissipated as heat/radiation during charge redistribution.)

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Electrostatic Potential and Capacitance

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