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Capacitance of dielectric-filled spherical shells

Short answer

Capacitance C = 4πε₀ k ab/(b−a) = 1.113×10⁻¹⁰ × 23.5 × 4.08×10⁻² ≈ 1.07×10⁻¹⁰ F (107 pF). Free charge q = CV = 1.07×10⁻¹⁰ × 73 ≈ 7.8×10⁻⁹ C (7.8 nC).

The space between two concentric conducting spherical shells of radii b = 1.70 cm and a = 1.20 cm is filled with a substance of dielectric constant k = 23.5. A potential difference V = 73 V is applied across the inner and outer shells. Determine (a) the capacitance of the device (b) the free charge q on the inner shell.

Given: inner radius a = 1.20 cm = 1.20 × 10⁻² m, outer radius b = 1.70 cm = 1.70 × 10⁻² m, k = 23.5, V = 73 V.

(a) Capacitance

For a spherical capacitor with vacuum, C₀ = 4πε0 ab/(b − a). With a dielectric filling the gap, C = k C₀.

  1. C = 4πε0 k ab/(b − a)
  2. ab = (1.20 × 10⁻²)(1.70 × 10⁻²) = 2.04 × 10⁻⁴ m²
  3. b − a = 0.50 × 10⁻² = 5.0 × 10⁻³ m
  4. ab/(b − a) = 2.04 × 10⁻⁴ / 5.0 × 10⁻³ = 4.08 × 10⁻² m
  5. 4πε0 = 1.113 × 10⁻¹⁰ F/m
  6. C = 1.113 × 10⁻¹⁰ × 23.5 × 4.08 × 10⁻² = 1.07 × 10⁻¹⁰ F

Answer (a): C ≈ 1.07 × 10⁻¹⁰ F = 107 pF

(b) Free charge on the inner shell

q = CV = 1.067 × 10⁻¹⁰ × 73 = 7.79 × 10⁻⁹ C

Answer (b): q ≈ 7.8 × 10⁻⁹ C = 7.8 nC (the free charge is on the inner shell; the outer shell carries −7.8 nC).

✓ Answer checked by Claude Opus 5.5 (high effort)

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