Question

In the Z-direction, the cross-section of a long conducting shell is a square with the side lengths of a in the x and y directions. The potential on the side y = 0 is

ϕ(x,y=0)=Ksin(2πx/a)
, and the other sides are grounded. Solve for the potential inside the shell. Starting from the general solution, proceed by using the boundary values on the grounded surfaces to derive the expression:
ϕ(x,y)=∑nAnsin(anπx)(e⁻ⁿπeⁿπʸ/a−eπe⁻ⁿπʸ/a)
and solve for the potential using the remaining boundary condition.

Answer 1

The potential inside the square conducting shell is determined using separation of variables, and Fourier series, applying the given boundary conditions to find the coefficients for the general solution.

Step-by-step explanation:

The potential inside a square conducting shell with varying potential on one side and grounded on the others can be solved using separation of variables and applying boundary conditions. The general solution for potential is a Fourier series summing over solutions that satisfy Laplace's equation. The boundary condition φ(x,y=0)=Ksin(2πx/a) allows us to set the coefficients for terms in the solution, obtaining the potential inside the shell.

The general solution given is φ(x,y) = ∑Ansin(anπx)(e⁻ⁿπ^nπy/a-eπ^nπy/a), where An coefficients need to be determined using the remaining boundary condition. By applying Fourier's trick, one multiplies both sides by sin(anπx) and integrates over the length of the side, solving for An and thus determining the specific potential distribution inside the shell.