(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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. Relevant equations

Laplace's equaton in rectangular coordinates in 2 dimensions

[tex] \frac{d^2 V}{dx^2} + \frac{d^2 V}{dy^2} = 0 [/tex]

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

[tex] V(x,y) = = X(x) Y(y) = (Ae^{kx} + Be^{-kx}) (C\cos ky + D\sin ky) [/tex]

using conditon 3, C=0

using condition 4, [itex] k = n\pi / a [/itex]

Now there is a symmetry such that V(0,y) = -V(a,y)

so then A = - B, and

[tex] X(x) = e^{kx} - e^{-kx} = 2 \sinh kx[/tex]

SO then the potential is

[tex] V(x,y) = \sum_{n=1}^{\infty} C_{n} \sinh\left(\frac{n\pi x}{a}\right) \sin\left(n\pi y}{a}\right) [/tex]

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

now V(0,y) = -V(a,y)

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

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

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

[tex] n\pi (\frac{x}{a} + q) [/tex]

when x = a, the argument is positive

[tex] n\pi (1+q) = n\pi[/tex]

[tex] n\pi (q) = n \pi [/tex]

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

[tex] n\pi (\frac{2x}{a} -1) [/tex]

but the textbook says there should be an additional factor of 1/2.. that is

[tex] \frac{n\pi}{2} (\frac{2x}{a} -1) [/tex]

how did they get that??

Thanks for help!

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# Homework Help: Laplace's equation

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