# 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

$V(x,y)=sum_{n=1}^{infty}C_n e^{-npi x/a}sin(npi y/a)$

The 4-th boundary condition is here $V(0,y)=(V_0/a)*x$ so that

$V(0,y)=sum_n C_nsin(npi y/a)=(V_0/a)*x$ $(1)$

Using the Fourier trick one has

$C_n=frac{2}{a}int_{-infty}^{infty}frac{V_0}{a}x*sin(frac{npi}{a}y)dy=frac{2V_0}{a^2}xint_0^asin(frac{npi}{a}y)dy$

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

$C_n=(2/a)int_0^a (V_0/a)y*sin(npi y/a)dy$

$C_1=2V_0/pi$, $C_2=-V_0/pi$, $C_3=2V_0/3pi$, $C_4=-V_0/2pi$, …

So that the potential inside the slot is

$V(x,y)=sum_{n=1}^{infty}(-1)^{n+1}*(2V_0/npi)*e^{-npi x/a}sin(npi y/a)$