---
title: "Potential between conducting planes"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:conducting-planes-potential-field"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/conducting-planes-potential-field
site: BCT Engineering
---

# Potential between conducting planes

Electromagnetism · Chapter 2: Electric field

- Asked in: 2076 Asoj, 7 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/conducting-planes-potential-field
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

The conducting planes 2x + 3y = 12 and 2x + 3y = 18 are at potentials of 100 V and 0, respectively. Let ε = ε0 and find: (a) V at P(5, 2, 6); (b) E at P(5, 2, 6).

## Short answer

Potential depends only on u=2x+3y; solving ∇²V=0 gives V=300−16.667u. At P(5,2,6), V=33.3 V. The uniform field is E=−∇V = 33.3 a_x + 50 a_y V/m.

## Answer

**Given:** Planes 2x + 3y = 12 at 100 V and 2x + 3y = 18 at 0 V, ε = ε₀. Find V and E at P(5, 2, 6).

**Method:** The potential depends only on the coordinate perpendicular to the planes. Define u = 2x + 3y. There is no charge between the planes, so Laplace's equation holds, and V is a function of u only.

#### Solve Laplace's equation

V = V(x, y) with u = 2x + 3y:

```text
∂²V/∂x² = 4 d²V/du²,    ∂²V/∂y² = 9 d²V/du²
∇²V = 13 d²V/du² = 0   →  V = C₁ u + C₂
```

#### Boundary conditions

```text
V = 100 at u = 12:  12 C₁ + C₂ = 100
V = 0   at u = 18:  18 C₁ + C₂ = 0
Subtract: 6 C₁ = −100  →  C₁ = −16.667,   C₂ = 300
```

```text
V = 300 − 16.667 (2x + 3y) = (100/6) (18 − 2x − 3y)
```

#### (a) V at P(5, 2, 6)

u = 2(5) + 3(2) = 16

```text
V = 300 − 16.667(16) = 300 − 266.67 = 33.3 V
```

(P lies between the planes: 12 < 16 < 18.)

#### (b) E at P

```text
E = −∇V = 16.667 (2 a_x + 3 a_y) = 33.3 a_x + 50 a_y   V/m
```

The field is uniform between the planes. Its magnitude is |E| = 16.667 √13 = 60.1 V/m, directed from the 100 V plane toward the 0 V plane, perpendicular to the planes.

**Answer: (a) V = 33.3 V; (b) E = 33.3 a_x + 50 a_y V/m**

## Check yourself

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

1. Why can V be taken as a function of u = 2x + 3y only?
   A. Because z is always zero
   B. The planes are parallel, so V varies only in the direction perpendicular to them
   C. Because E is zero between the planes
   D. Because the planes are at equal potential
   Answer: B. Equipotential planes of constant u mean V depends only on u. (Key point: Variable choice)

2. Laplace's equation with V(u) gives:
   A. ∇²V = 13 d²V/du² = 0, so V = C₁u + C₂
   B. ∇²V = 5 d²V/du² = 0, so V = C₁u² + C₂
   C. ∇²V = 13 dV/du = 0, so V = C₂ e^u
   D. ∇²V = 13 d²V/du² = 0, so V = C₁ ln u + C₂
   Answer: A. ∂²V/∂x² = 4V'' and ∂²V/∂y² = 9V'', giving 13V'' = 0 and V linear in u. (Key point: Laplace solution)

3. At P(5, 2, 6), the potential V is:
   A. 66.7 V
   B. 50 V
   C. 16 V
   D. 33.3 V
   Answer: D. u = 16, and V = 300 - 16.667(16) = 33.3 V. (Key point: Potential at P)

4. E at P is 33.3 a_x + 50 a_y V/m. Which statement is correct?
   A. It is parallel to the planes
   B. It varies with z
   C. It is uniform and perpendicular to the planes, from the 100 V plane to the 0 V plane
   D. It points from the 0 V plane to the 100 V plane
   Answer: C. E = -∇V = 16.667(2a_x + 3a_y), normal to the planes and pointing toward lower potential. (Key point: Field direction)

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