Simplify the expression -- Electric field in a capacitor

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
skrat
Messages
740
Reaction score
8

Homework Statement


Between two metal plates a metal strip is inserted as shown on the figure.
Captureh.PNG

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.

Homework Equations

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? o_O

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:
Physics news on Phys.org
skrat said:

Homework Statement


Simplifly ##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)## for ##x>>a##.

Homework Equations

The Attempt at a Solution



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? o_O

ps: If you need the whole problem, just let me know, but it shouldn't play any role for this...

Consider the expression (1 - (-1)n). What happens when n is even? When n is odd?
 
SteamKing said:
Consider the expression (1 - (-1)n). What happens when n is even? When n is odd?
How is this related to ##x>>a##?

ps: SteamKing, I edited the post and added some more questions.
 
skrat said:
How is this related to ##x>>a##?

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

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.
 
SteamKing said:
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.

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##.