# Laplace's equation

1. Homework Statement
Consider a square in the XY plane with corners at (0,0, (a,0), (a,a,) and (0,a). There is no charge nor matter inside the square. The sides perpendicular to the Y axis have potential zero. The side at x=a has constant potentail V0. The side at x=0 has potentail -V0. Find V(x,y) at all points inside the square.

2. Homework Equations
Laplace's equaton in rectangular coordinates in 2 dimensions

$$\frac{d^2 V}{dx^2} + \frac{d^2 V}{dy^2} = 0$$

3. The Attempt at a Solution
Boundary conditions are
1. V=0, y=a
2. V=0, y=0
3. V=V0, x= a
4. V=-V0, x=0

By separation of variables
$$V(x,y) = = X(x) Y(y) = (Ae^{kx} + Be^{-kx}) (C\cos ky + D\sin ky)$$

using conditon 3, C=0
using condition 4, $k = n\pi / a$
Now there is a symmetry such that V(0,y) = -V(a,y)

so then A = - B, and
$$X(x) = e^{kx} - e^{-kx} = 2 \sinh kx$$

SO then the potential is
$$V(x,y) = \sum_{n=1}^{\infty} C_{n} \sinh\left(\frac{n\pi x}{a}\right) \sin\left(n\pi y}{a}\right)$$

The Cn coefficients are simply $4V_{0}/n \pi$ if n is odd otherwise if n is even Cn is zero

now V(0,y) = -V(a,y)
$$[tex] V(x,y) = \sum_{n=1}^{\infty} C_{n} \sinh\left(n\pi \right) \sin\left(n\pi y}{a}\right) = V_{0}$$
$$V(0,y) = \sum_{n=1}^{\infty} C_{n} \sinh(0) \sin\left(n\pi y}{a}\right) =-V_{0}$$

but sinh (0) = 0 ... the argument for the sinh must be wrong

i know sinh is odd function so somehow the argumet of sinh when x= 0 must be equal to the negative of argument of sinh when x = a

do i simply have to guess this?? Or is there a more step by step approach??

i was guessing the argument should be something like this

$$n\pi (\frac{x}{a} + q)$$
when x = a, the argument is positive
$$n\pi (1+q) = n\pi$$
$$n\pi (q) = n \pi$$

q should be -1 but then in the first case the argumetn becomes zero. So we put a 2 in front of that fraction

$$n\pi (\frac{2x}{a} -1)$$
but the textbook says there should be an additional factor of 1/2.. that is

$$\frac{n\pi}{2} (\frac{2x}{a} -1)$$
how did they get that??

Thanks for help!

Last edited:

## Answers and Replies

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LeonhardEuler
Gold Member
so then A = - B, and
$$X(x) = e^{kx} - e^{-kx} = 2 \sinh kx$$
Look at this again. That symetry argument applies for V(-a,y)=-V(a,y). Is that what you have here?

Look at this again. That symetry argument applies for V(-a,y)=-V(a,y). Is that what you have here?

ahh no i have -V(0,y) = V(a,y)

so then the argment of the exponential has to change as well??
such that at 0 the argument turns to -a??

something along these lines??

LeonhardEuler
Gold Member
ahh no i have -V(0,y) = V(a,y)

so then the argment of the exponential has to change as well??
such that at 0 the argument turns to -a??

something along these lines??
No, I would abandon the symetry argument altogether and deal with the boundary conditions individually.

No, I would abandon the symetry argument altogether and deal with the boundary conditions individually.
ok abandoning the symmetry argument

then we get tow equation

$$A + B = -V_{0}$$
$$Ae^{n\pi} + Be^{-n\pi} = V_{0}$$

iget

$$A = V_{0} \frac{1 + e^{-n\pi}}{2\sinh n\pi}$$
$$B= -V_{0} \frac{1+ e^{n\pi}}{2\sinh n\pi}$$

$$X(x) = V_{0} \frac{1 + e^{-n\pi}}{2\sinh n\pi}-V_{0} \frac{1+ e^{n\pi}}{2\sinh n\pi} = \frac{V_{0}e^{\frac{n\pi x}{a}}}{2\sinh n\pi} (e^{-n\pi} - e^{n\pi})$$

but
$$e^{-n\pi} - e^{n\pi} = -2\sinh n\pi$$
so
$$X(x) = -\frac{V_{0}}{2} e^{\frac{n\pi x}{a}}$$

i get a feeling i did something wrong somehwere but i am unable to track the error...

Astronuc
Staff Emeritus
1. V=0, y=a
2. V=0, y=0

The yBC could imply no dependence, i.e. the function of y=1, which means the solution is a function of x only. The BC also imply an even function about a/2.

3. V=V0, x= a
4. V=-V0, x=0

The xBC imply an odd function about a/2, e.g. A sin kx or A sinh kx (or Aekx - A e-kx. There are two unknowns A and k which can be solved using the BC.

1. V=0, y=a
2. V=0, y=0

The yBC could imply no dependence, i.e. the function of y=1, which means the solution is a function of x only. The BC also imply an even function about a/2.

3. V=V0, x= a
4. V=-V0, x=0

The xBC imply an odd function about a/2, e.g. A sin kx or A sinh kx (or Aekx - A e-kx. There are two unknowns A and k which can be solved using the BC.
i got it now... my solution wold have worked if the square was centered about the origin. But in this case we needed to 'shift' the solution suhc that it conforms with our symmetry