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Parallel plate capacitor with mica

Short answer

Plate separation d = kε₀A/C ≈ 4.78×10⁻⁹ m. (a) E = V/d ≈ 1.05×10¹⁰ V/m. (b) Free charge q = CV = 5×10⁻³ C (5 mC). (c) Induced surface charge q′ = q(1−1/k) ≈ 4.07×10⁻³ C (4.07 mC).

A parallel plate capacitor has a capacitance of 100 μF, a plate area of 100 cm² and a mica dielectric. At 50 volts potential difference calculate (a) E in the mica (b) the free charge on the plates and (c) the induced surface charge. [Dielectric constant for mica, k = 5.4]

Given: C = 100 μF = 10⁻⁴ F, A = 100 cm² = 10⁻² m², V = 50 V, k = 5.4, ε0 = 8.854 × 10⁻¹² F/m. (The numbers as given make the plate separation extremely small; we simply use them as stated. In the textbook version of this problem C = 100 pF, which gives E ≈ 1.05 × 10⁴ V/m, q = 5 nC and q' ≈ 4.07 nC by the same steps.)

Plate separation

C = kε0A/d → d = kε0A/C = (5.4 × 8.854 × 10⁻¹² × 10⁻²)/10⁻⁴ = 4.78 × 10⁻⁹ m

(a) Electric field in the mica

E = V/d = 50 / 4.78 × 10⁻⁹ = 1.05 × 10¹⁰ V/m

Answer (a): E ≈ 1.05 × 10¹⁰ V/m

(b) Free charge on the plates

q = CV = 10⁻⁴ × 50 = 5 × 10⁻³ C

Answer (b): q = 5 mC

(c) Induced surface charge on the mica

The dielectric reduces the field, so q' = q(1 − 1/k):

q' = 5 × 10⁻³ (1 − 1/5.4) = 5 × 10⁻³ × 0.8148 = 4.07 × 10⁻³ C

(Check: E = q/(kε0A) = 5 × 10⁻³ / (5.4 × 8.854 × 10⁻¹² × 10⁻²) = 1.05 × 10¹⁰ V/m, the same.)

Answer (c): q' ≈ 4.07 mC, opposite in sign to the adjacent plate's free charge.

✓ Answer checked by Claude Opus 5.5 (high effort)

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