# Simplify the expression -- Electric field in a capacitor

1. Nov 11, 2014

### skrat

1. The problem statement, all variables and given/known data
Between two metal plates a metal strip is inserted as shown on the figure.
a) Calculate the electric potential anywhere inside the capacitor.
b) Simplifly $U(x,y)$ for $x>>a$.
c) Calculate the electric field in the middle of the capactior.

2. Relevant equations

3. The attempt at a solution

a) Ok, since we have no other charges in the capacitor, than $\nabla ^2 U=0$. In kartezian coordinates this means that $U(x,y)=\sum _m (A_me^{-k_mx}+B_me^{k_mx})(Csin(k_my)+Dcos(k_my))$.

Using boundary conditions
$U(0,y)=U_0$ and
$U(x,0)=0$ and
$U(x,a)=0$ we find out that

$U(x,y)=\sum _{n=1} ^{\infty}\frac{2U_0}{n\pi }(1-(-1)^n)e^{-\frac{n\pi}{a}x}\sin(\frac{n\pi }{a}y)$

b) The result is that out of the sum only $n=1$ survives, therefore $U(x,y)=\frac{4U_0}{\pi }e^{-\frac{\pi}{a}x}\sin(\frac{\pi }{a}y)$.

I am having some troubles understanding this. Could anybody please try to explain how do I get this result?

c)

$U(x,a/2)=\sum _{n=1} ^{\infty}\frac{2U_0}{n\pi }(1-(-1)^n)e^{-\frac{n\pi}{a}x}\sin(\frac{n\pi }{2})$

$\vec E=-\nabla U(x,a/2)=\sum _{n=1} ^{\infty}\frac{2U_0}{a}(1-(-1)^n)\sin(\frac{n\pi }{2})e^{-\frac{n\pi}{a}x}$

I think I should do something more with that last equation but I really don't know what or how.. :/

Last edited: Nov 11, 2014
2. Nov 11, 2014

### SteamKing

Staff Emeritus
Consider the expression (1 - (-1)n). What happens when n is even? When n is odd?

3. Nov 11, 2014

### skrat

How is this related to $x>>a$?

ps: SteamKing, I edited the post and added some more questions.

4. Nov 11, 2014

### SteamKing

Staff Emeritus
Well, if you had examined the expression, you would have seen that about half the terms drop out, due to the expression being 0 for n even.

5. Nov 12, 2014

### skrat

Yes, I agree. But this still doesn't explain why only $n=1$ is ok? What happens with $n=3$?
I apologize but I still can't see how is this related to $x>>a$.