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

# Capacitance of spherical capacitor

Electromagnetism · Chapter 2: Electric field

- Asked in: 2075 Chaitra, 5 marks; 2075 Asoj, 6 marks (asked 2 times)
- Video: [Spherical Capacitor: Electric Field and Capacitance Explained](https://www.youtube.com/watch?v=w7NsDz8u1Sc) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/spherical-capacitor-laplace
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Find the capacitance of a spherical capacitor using Laplace's equation (assume the potential V is a function of r only and solve by the integration method).

## Short answer

Solving Laplace's equation in spherical coordinates for V(r) between concentric spheres (radii a, b, potentials V₀, 0) yields V = C₁/r + C₂. Applying boundary conditions gives C₁ = V₀ ab/(b−a). Electric field E = (V₀ ab/(b−a))(1/r²) a_r, charge Q = 4πε ab V₀/(b−a), so capacitance C = Q/V₀ = 4πε ab/(b−a) F.

## Answer

**Aim:** Capacitance C = Q/V₀ of two concentric spherical conductors: inner radius a at potential V₀, outer radius b at potential 0, with a dielectric of permittivity ε between them.

```text
     ╭─────────╮
    ╱  ╭─────╮  ╲      inner sphere: radius a, V = V₀
   │  │  ●─a │   │     outer sphere: radius b, V = 0
    ╲  ╰─────╯  ╱      region a < r < b: ρ_v = 0
     ╰─────b───╯
```

#### Step 1: Laplace's equation

The region between the spheres has no free charge, so ∇²V = 0. By spherical symmetry V depends only on r:

```text
(1/r²) d/dr ( r² dV/dr ) = 0
```

#### Step 2: Integrate twice

```text
r² dV/dr = A                    (first integration; A is a constant)
dV/dr = A / r²
V = −A / r + B                  (second integration)
```

Write the constants as C₁ = −A and C₂ = B: V = C₁/r + C₂.

#### Step 3: Boundary conditions

```text
V(r = b) = 0:      C₁/b + C₂ = 0     →  C₂ = −C₁/b
V(r = a) = V₀:     C₁/a + C₂ = V₀    →  C₁ (1/a − 1/b) = V₀

C₁ = V₀ a b / (b − a)
```

```text
V(r) = V₀ ( ab / (b − a) ) ( 1/r − 1/b )
```

#### Step 4: Electric field and charge

```text
E = −∇V = −(dV/dr) a_r = ( V₀ ab / (b − a) ) (1/r²) a_r
```

Surface charge density on the inner sphere (r = a): ρ_s = D_n = ε E_r(a) = ε V₀ b / ( a (b − a) ).

```text
Q = ρ_s (4πa²) = 4π ε a b V₀ / (b − a)
```

#### Step 5: Capacitance

```text
C = Q / V₀ = 4π ε a b / (b − a)      farads
```

**Result: C = 4πε ab / (b − a)**, with ε = ε₀ε_r.

**Special case:** For b → ∞ (an isolated sphere), C = 4πεa.

## Check yourself

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

1. For the spherical capacitor with V a function of r only and no charge between the spheres, which equation is solved?
   A. d/dr (dV/dr) = 0
   B. (1/r) d/dr (r dV/dr) = 0
   C. (1/r²) d/dr (r² dV/dr) = 0
   D. d²V/dr² = -ρ_v/ε
   Answer: C. In spherical coordinates with only r dependence, ∇²V = (1/r²) d/dr(r² dV/dr) = 0. (Key point: Laplace's equation form)

2. Integrating twice gives which general potential?
   A. V = C₁ ln r + C₂
   B. V = C₁/r + C₂
   C. V = C₁ r + C₂
   D. V = C₁/r² + C₂
   Answer: B. r² dV/dr = A gives dV/dr = A/r², so V = -A/r + B, i.e. C₁/r + C₂. (Key point: General solution)

3. Which boundary conditions are used to find the constants?
   A. V = V₀ at r = a and V = 0 at r = b
   B. V = 0 at r = a and V = V₀ at r = b
   C. E = 0 at r = a and V = V₀ at r = b
   D. V = V₀ at r = 0 and V = 0 at r = ∞ only
   Answer: A. The inner sphere (radius a) is at V₀ and the outer sphere (radius b) is at 0. (Key point: Boundary conditions)

4. What is the final capacitance of the spherical capacitor, and its isolated-sphere limit?
   A. C = 2πεL/ln(b/a); for b → ∞, C = 0
   B. C = 4πε(b - a)/ab; for b → ∞, C = 4πεa
   C. C = 4πε ab/(b - a); for b → ∞, C = infinity
   D. C = 4πε ab/(b - a); for b → ∞, C = 4πεa
   Answer: D. C = Q/V₀ = 4πεab/(b - a), and as b → ∞ this tends to 4πεa. (Key point: Capacitance result)

## Related in Electromagnetism

- [Capacitance of coaxial capacitor](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Acoax-capacitance-laplace-cylindrical) — Electromagnetism, chapter 2, asked 2×
- [Capacitance of parallel plate capacitor](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Apoisson-parallel-plate-capacitance) — Electromagnetism, chapter 2
- [2D Laplace potential distribution](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Alaplace-2d-potential-distribution) — Electromagnetism, chapter 2
- [Scalar potential, Laplace, magnetic flux](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Ascalar-magnetic-potential-laplace) — Electromagnetism, chapter 3

## Related in other subjects

- [General capacitance method, spherical capacitor](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acapacitance-general-method-spherical) — Engineering Physics, chapter 7
- [Capacitance of dielectric-filled spherical shells](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aspherical-shells-dielectric-capacitance) — Engineering Physics, chapter 7
- [Capacitance definition and charge equation](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Acapacitance-charge-definition) — Basic Electrical Engineering, chapter 4
- [Capacitance of cylindrical capacitor](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acapacitance-general-method-cylindrical) — Engineering Physics, chapter 7, asked 2×