Potential from Laplace Equation
Solve for the electrostatic potential, $V (x, y)$, in the region $x \geq 0$, $0 \leq y \leq a$, defined by boundary conditions
$V(x, 0) = 0$
$V(x, a) = 0$
$V(0, y) = V_0$
$V (\infty , y) = 0$ .
The charge density vanishes everywhere inside the boundaries.
The 2D Laplace equation for potential with no charge is
$∆V(x,y)=0$ or $(d^2 V)/(dx^2 )+(d^2 V)/(dy^2 )=0$
With the boundary conditions
$V(x,0)=0$ (1)
$V(x,a)=0$ (2)
$V(0,y)=V_0$ (3)
$V(∞,y)=0$ (4)
The solution of the above Laplace equation can be written as
$V(x,y)=X(x)*Y(y)$ so that
$Y* \frac {d^2 X}{dx^2}+X*\frac{d^2 Y}{dy^2}=0$ or
$(1/X)*\frac{d^2 X}{dx^2 }+(1/Y)*\frac{d^2 Y}{dy^2}=0$
Which is satisfied if each term in the equation is the same constant $C$ (just with changed sign): $C-C=0$
$1/X*\frac{d^2 X}{dx^2}=Constant$ and $1/Y*\frac{d^2 Y}{dy^2}=-Constant$
Having as solutions
$X(x)=Ae^{ikx}+Be^{-ikx}$ and $Y(y)=C\sin (ky)+D\cos(ky)$ where $k^2=Constant$
So that
$V(x,y)=(Ae^{ikx}+Be^{-ikx} )[C\sin(ky)+D\cos(ky)]$
From the boundary conditions
(4) $x→∞$ means $V=0$ so $X(x)=B*e^{-ikx}$
$V(x,y)=e^{-ikx} [C\sin(ky)+D cos(ky)]$ with coeffcient B being absorbed into C and D〗 〗
(1) $y=0$ means $V=0$ so $D=0$ and $V(x,y)= C*e^{-ikx}*sin(ky)$
(2) $y=0$ means $V=0$ so that $\sin(ka)=0$ or $k=nπ/a$
So that finally
$V(x,y)=C*e^{-ikx}*\sin(nπ/a*y)$
V(x,y) is the infinite sum of all these components for all possible n integer
$V(x,y)=∑_(n=1)^∞[C_n*e^{i(nπ/a)*x}*sin((nπ/a)*y)]$
Last boundary condition is
(3) $V(0,y)=V_0$ so that $∑_n [C_n*\sin((nπ/a)*y)]=V_0]$
The coefficients $C_n$ can be found using the Fourier trick
$\int_0^a [sin((nπ/a)*y)*sin((mπ/a)* y)dy]=\left\{\begin{matrix}
0 & \text {for }n\neq m\\
a/2 & \text {for }n=m
\end{matrix}\right.$
So that
$C_n=2/a*\int_0^a [V_0*sin((nπ/a)*y)*dy] ==\left\{\begin{matrix}
(2V_0/a)(2a/n\pi) & \text {for n odd}\\
0 & \text {for n even}
\end{matrix}\right.$
So that finally
$V(x,y)=∑_{n-odd}[\frac {4V_0}{n}*sin((nπ/a)*y)]$