Mathematical Tools for Physics

10—Partial Differential Equations 301

−L 0 L 2 L x

z

y V^0

Arrange a set of conducting strips in thex-yplane and with insulation between them so that they don’t
quite touch each other. Now apply voltageV 0 on every other one so that the potentials are alternately zero and
V 0. This sets the potential in thez= 0plane to be independent ofyand

z= 0 : V(x,y) =

{

V 0 ( 0 < x < L)

0 (L < x < 2 L)

V(x+ 2L,y) =V(x,y), allx,y (32)

What is then the potential above the plane,z > 0? Above the planeV satisfies Laplace’s equation,

∇^2 V=

∂^2 V

∂x^2

+

∂^2 V

∂y^2

+

∂^2 V

∂z^2

= 0 (33)

The potential is independent ofyin the plane, so it will be independent ofyeverywhere. Separate variables in
the remaining coordinates.

V(x,z) =f(x)g(z) =⇒

d^2 f

dx^2

g+f

d^2 g

dz^2

= 0 =⇒

1

f

d^2 f

dx^2

+

1

g

d^2 g

dz^2

= 0

This is separated as a function ofxplus a function ofy, so the terms are constants.

1

f

d^2 f

dx^2

=−α^2 ,

1

g

d^2 g

dz^2

= +α^2 (34)

I’ve chosen the separation constant in this form because the boundary condition is periodic inx, and that implies
that I’ll want oscillating functions there, not exponentials.

f(x) =eiαx and f(x+ 2L) =f(x)

=⇒e^2 Liα= 1, or 2 Lα= 2nπ, n= 0,± 1 ,± 2 ,...