# Infinite Slot Potential

What is the potential of the infinite slot from Griffiths 3.3, if you are given a varying potential $V(0,y)=V_0x/a$ as a boundary condition at $x=0$

For the given arrangement of example 3.3, the general solution of the potential that satisfies the 3 zero boundary conditions ($V=0$ in $y=0$, in $y=a$ and when $\to \infty$) is
$V(x,y)=\sum_{n=1}^{\infty}C_n e^{-n\pi x/a}\sin(n\pi y/a)$
The 4-th boundary condition is here $V(0,y)=(V_0/a)*x$ so that
$V(0,y)=\sum_n C_n\sin(n\pi y/a)=(V_0/a)*x$           $(1)$
Using the Fourier trick one has
$C_n=\frac{2}{a}\int_{-\infty}^{\infty}\frac{V_0}{a}x*\sin(\frac{n\pi}{a}y)dy=\frac{2V_0}{a^2}x\int_0^a\sin(\frac{n\pi}{a}y)dy$

with $C_1=\frac{2V_0}{a^2}*\frac{2a}{\pi}x=\frac{4V_0}{\pi a}x$  and the rest of $C_n=0$
so that $V(x,y)=(4V_0/\pi a)*x e^{\pi x/a}*\sin(\pi y/a)$

If the 4th boundary condition was $V(0,y)=(V_0/a)y$  the coefficients of the expansion were
$C_n=(2/a)\int_0^a (V_0/a)y*\sin(n\pi y/a)dy$
$C_1=2V_0/\pi$,  $C_2=-V_0/\pi$, $C_3=2V_0/3\pi$, $C_4=-V_0/2\pi$, …

So that the potential inside the slot is
$V(x,y)=\sum_{n=1}^{\infty}(-1)^{n+1}*(2V_0/n\pi)*e^{-n\pi x/a}\sin(n\pi y/a)$