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Potential between conducting planes

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.

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).

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:

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

Boundary conditions

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

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

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

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

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

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

(b) E at P

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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