---
title: "Parallel plate capacitor with mica"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:mica-capacitor-field-free-charge"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/mica-capacitor-field-free-charge
site: BCT Engineering
---

# Parallel plate capacitor with mica

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2073 Chaitra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/mica-capacitor-field-free-charge
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

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]

## 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).

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the free charge on the plates found?
   A. q = C/V = 2 μC
   B. q = kCV = 27 mC
   C. q = CV = 10^-4 x 50 = 5 mC
   D. q = CV/k ≈ 0.93 mC
   Answer: C. Free charge is the charge delivered to the plates, q = CV. (Key point: Free charge)

2. How is E in the mica found from the given data?
   A. E = V x d
   B. d = kε0A/C from C = kε0A/d, then E = V/d
   C. E = q/(ε0 A) with no k
   D. E = CV/A
   Answer: B. d = 4.78 x 10^-9 m gives E = V/d ≈ 1.05 x 10^10 V/m. (Key point: Plate separation and E)

3. What is the induced (bound) surface charge q'?
   A. q(1 - 1/k) ≈ 4.07 mC
   B. q/k ≈ 0.93 mC
   C. q(1 + 1/k) ≈ 5.9 mC
   D. kq ≈ 27 mC
   Answer: A. The dielectric reduces the field so that q - q' = q/k. (Key point: Induced charge)

4. What is the sign of the induced charge on a mica face?
   A. Same as the adjacent plate's free charge
   B. Always zero
   C. Always larger than the free charge
   D. Opposite to the adjacent plate's free charge, partly neutralising it
   Answer: D. Bound charge on the face next to a plate is opposite to the plate's free charge. (Key point: Sign of induced charge)

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