gysush
Mar8-11, 07:38 PM
We solve Poisson's equation in cartesian coords for a region bounded by planes forming a box. Some of the planes are grounded. The lengths of the box are L1, L2, L3. There is no charge distribution.
Let a=mPi/L2
b=nPi/L3
The solutions goes like: sinh(sqrt(a^2 + b^2)*x)sin(a*y)sin(b*z)
with our coefficients going like: 1/(m*n)*1/sinh(sqrt(a^2 + b^2)*L1) (neglecting factors of Pi and constants)
The solution is valid only for odd n,m from 1 to infinity.
I want to show that the avg value of the potential over all the sides is equal to the potential at the center:
V(L1/2,L2/2,L3/2) = prescribed equation asked to show.
So I evaluate all my terms at the center. I have some cancellation of the sinh terms because sinh(2x)=2*sinh(x)*cosh(x)
I make a change of variables from m,n to 2k-1,2j-1 so that the sum goes from 1 to infinity for all positive integers and thus allowing me to change my two sin terms into (-1)^j+k
So what I am trying to sum is this:
Sum(j:1,00)Sum(k:1,00) (-1)^j+k *1/[(2k-1)(2j-1)]1/cosh(sqrt(a^2 + b^2)*L1)
a=(2j-1)Pi/L2
b=(2k-1)Pi/L3
Kinda stumped as to any analytical route without trying to use software.
Let a=mPi/L2
b=nPi/L3
The solutions goes like: sinh(sqrt(a^2 + b^2)*x)sin(a*y)sin(b*z)
with our coefficients going like: 1/(m*n)*1/sinh(sqrt(a^2 + b^2)*L1) (neglecting factors of Pi and constants)
The solution is valid only for odd n,m from 1 to infinity.
I want to show that the avg value of the potential over all the sides is equal to the potential at the center:
V(L1/2,L2/2,L3/2) = prescribed equation asked to show.
So I evaluate all my terms at the center. I have some cancellation of the sinh terms because sinh(2x)=2*sinh(x)*cosh(x)
I make a change of variables from m,n to 2k-1,2j-1 so that the sum goes from 1 to infinity for all positive integers and thus allowing me to change my two sin terms into (-1)^j+k
So what I am trying to sum is this:
Sum(j:1,00)Sum(k:1,00) (-1)^j+k *1/[(2k-1)(2j-1)]1/cosh(sqrt(a^2 + b^2)*L1)
a=(2j-1)Pi/L2
b=(2k-1)Pi/L3
Kinda stumped as to any analytical route without trying to use software.