---
title: "Capacitance of parallel plate capacitor"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:poisson-parallel-plate-capacitance"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/poisson-parallel-plate-capacitance
site: BCT Engineering
---

# Capacitance of parallel plate capacitor

Electromagnetism · Chapter 2: Electric field

- Asked in: 2074 Asoj, 5 marks
- Video: [Parallel Plate Capacitor: Electric Field and Capacitance Explained | Electrostatics](https://www.youtube.com/watch?v=-vwo43a-7G4) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/poisson-parallel-plate-capacitance
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

By solving Laplace's equation, find the capacitance of a parallel plate capacitor with potential difference between the plates equal to V0.

## Short answer

Solving Laplace's equation d²V/dy²=0 with V(0)=0, V(d)=V₀ yields V = V₀ y/d. Then E = −V₀/d a_y, D = εE, surface charge ρ_s = ε V₀/d, so Q = ε V₀ S/d. Capacitance C = ε S/d.

## Answer

**Setup.** Two large parallel plates of area S are separated by distance d, with the dielectric of permittivity ε between them. Fringing is neglected, so the potential V depends only on y. Let the lower plate (y = 0) be at V = 0 and the upper plate (y = d) at V = V₀.

```text
 y = d  ======================  V = V0
          ↓  E = −(V0/d) a_y
 y = 0  ======================  V = 0
```

#### Solving Laplace's equation
Between the plates there is no free charge, so ∇²V = 0, which reduces to

```text
d²V/dy² = 0
dV/dy   = A
V       = A y + B
```
Boundary conditions:
- V = 0 at y = 0 gives B = 0.
- V = V₀ at y = d gives A = V₀/d.

```text
V = V₀ y / d
```

#### Field, charge and capacitance
```text
E = −∇V = −(V₀/d) a_y
D = εE  = −(εV₀/d) a_y
```
The field leaves the plate at the higher potential, so the upper plate (V₀) carries +Q and the lower plate carries −Q. The magnitude of the surface charge density is

```text
|ρ_s| = εV₀/d
Q = ρ_s S = εV₀S/d
```

Capacitance C = Q/V₀:

```text
C = εS/d = ε₀ε_r S/d   farads
```

**Answer: C = εS/d**, independent of V₀. Energy stored is W = ½CV₀² = εSV₀²/(2d).

## Check yourself

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

1. With V depending only on y, Laplace's equation becomes:
   A. dV/dy = 0
   B. d²V/dy² = −ρ_v/ε
   C. d²V/dy² = 0
   D. d/dy(y dV/dy) = 0
   Answer: C. Between the plates ∇²V reduces to d²V/dy² = 0. (Key point: Reduced Laplace)

2. With V = 0 at y = 0 and V₀ at y = d:
   A. V = V₀ y²/d²
   B. V = V₀ y/d
   C. V = V₀ ln(y)/ln(d)
   D. V = V₀ d/y
   Answer: B. Linear solution V = Ay + B with B = 0 and A = V₀/d. (Key point: Potential)

3. The magnitude of surface charge density on the plates is:
   A. εV₀/d
   B. V₀/(εd)
   C. εV₀ d
   D. ε/(V₀ d)
   Answer: A. ρ_s = D_n = εE = εV₀/d. (Key point: Charge density)

4. The capacitance is:
   A. C = εd/S
   B. C = εSV₀/d
   C. C = S/(εd)
   D. C = εS/d, independent of V₀
   Answer: D. Q = ρ_s S = εSV₀/d, C = Q/V₀ = εS/d. (Key point: Capacitance)

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