---
title: "Capacitance of coaxial capacitor"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:coax-capacitance-laplace-cylindrical"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/coax-capacitance-laplace-cylindrical
site: BCT Engineering
---

# Capacitance of coaxial capacitor

Electromagnetism · Chapter 2: Electric field

- Asked in: 2072 Chaitra, 6 marks; 2070 Chaitra, 5 marks (asked 2 times)
- Video: [Co axial Capacitor using Laplace's](https://www.youtube.com/watch?v=miPVuwuJr88) — Fareen's Video Hub
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/coax-capacitance-laplace-cylindrical
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Assuming that the potential V in the cylindrical coordinate system is a function of ρ only, solve Laplace's equation and derive the expression for the capacitance of a coaxial capacitor of length L using the same solution of V. Assume the inner conductor of radius a is at potential V0 with respect to the conductor of radius b.

## Short answer

With V(ρ) only, Laplace's equation (1/ρ)d/dρ(ρ dV/dρ)=0 integrates to V = A ln ρ + B. Using V(a)=V₀, V(b)=0 gives V = V₀ ln(b/ρ)/ln(b/a). Electric field E = (V₀/(ρ ln(b/a))) a_ρ. Surface charge yields Q = 2πεL V₀/ln(b/a), thus capacitance C = 2πεL/ln(b/a) F.

## Answer

**Aim:** Solve Laplace's equation for the region between coaxial cylinders (inner radius a at potential V₀, outer radius b at potential 0, length L, dielectric ε) and find the capacitance.

```text
        ╭──────────────╮
       ╱  ╭──────────╮  ╲    inner: ρ = a, V = V₀
      │  │   ●──a    │  │    outer: ρ = b, V = 0
       ╲  ╰──────────╯  ╱    length L (edge effects ignored)
        ╰──────b───────╯
```

#### Step 1: Laplace's equation

There is no charge between the cylinders, so ∇²V = 0. V depends only on ρ, so:

```text
(1/ρ) d/dρ ( ρ dV/dρ ) = 0
```

#### Step 2: Integrate twice

```text
ρ dV/dρ = A          →   dV/dρ = A/ρ
V = A ln ρ + B
```

#### Step 3: Boundary conditions

```text
V(b) = 0:    A ln b + B = 0    →  B = −A ln b
V(a) = V₀:   A ln a − A ln b = V₀   →  A = −V₀ / ln(b/a)
```

```text
V(ρ) = V₀ ln(b/ρ) / ln(b/a)
```

#### Step 4: Field and charge

```text
E = −dV/dρ a_ρ = ( V₀ / (ρ ln(b/a)) ) a_ρ
```

On the inner conductor (ρ = a): ρ_s = εE_ρ(a) = ε V₀ / (a ln(b/a)).

```text
Q = ρ_s × (2πaL) = 2π ε L V₀ / ln(b/a)
```

#### Step 5: Capacitance

```text
C = Q / V₀ = 2π ε L / ln(b/a)      farads
```

**Result: C = 2πεL / ln(b/a)**, or per unit length C/L = 2πε/ln(b/a) F/m.

## Check yourself

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

1. For V a function of ρ only, Laplace's equation in cylindrical coordinates becomes:
   A. d/dρ (ρ² dV/dρ) = 0
   B. d²V/dρ² = 0
   C. (1/ρ²) d/dρ (ρ dV/dρ) = 0
   D. (1/ρ) d/dρ (ρ dV/dρ) = 0
   Answer: D. Only the radial term remains: (1/ρ) d/dρ(ρ dV/dρ) = 0. (Key point: Laplace form)

2. The general solution after integrating twice is:
   A. V = A/ρ + B
   B. V = A ρ + B
   C. V = A ln ρ + B
   D. V = A ρ² + B
   Answer: C. ρ dV/dρ = A gives dV/dρ = A/ρ, so V = A ln ρ + B. (Key point: General solution)

3. With V(a) = V₀ and V(b) = 0, the field between the cylinders is:
   A. E = V₀ / (ln(b/a)) a_ρ, uniform
   B. E = V₀ / (ρ ln(b/a)) a_ρ, pointing outward
   C. E = V₀ ρ / ln(b/a) a_ρ
   D. E = V₀ / (ρ² ln(b/a)) a_ρ
   Answer: B. E = -dV/dρ with V = V₀ ln(b/ρ)/ln(b/a) gives V₀/(ρ ln(b/a)). (Key point: Potential and field)

4. The capacitance of the coaxial capacitor of length L is:
   A. C = 2πεL / ln(b/a)
   B. C = 2πεL ln(b/a)
   C. C = 4πε L / ln(b/a)
   D. C = πεL / ln(b/a)
   Answer: A. Q = ρ_s(2πaL) = 2πεLV₀/ln(b/a), so C = Q/V₀ = 2πεL/ln(b/a). (Key point: Capacitance)

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