Question

The walls of the box are infinite and you have a potential equal to zero, but the base that is on the x axis has a V(x,y)=V

Find the function of the potential V (x, y) with the boundary conditions

in the book uses a Fourier trick

Answer 1

The Laplace equation is
  
The Laplace equation is nabla^2 V(x, y) = 0 partial differential^2 V/partial differential x^2 + partial differential^2 V/partial differential y^2 = 0 We apply the separation of variables V(x, y) = X(x) Y(y) And so, the equation reduces to 1/X d^2 X/dx^2 + 1/Y d^2 Y/dy^2 = 0 1/X d^2 X/dx^2 = -1/Y d^2Y/dy^2 = -k^2 X(x) = A cos (kx) + B sin (kx) And Y(y) = Ce^ky + D^-ky And given the boundary conditions V(0, y) = 0, V(a, y) = 0, V(x, 0) = V_0, V(x, infinity) = 0 And so, the first condition is X(0) = A = 0 And so, X(x) = B sin (kx) And the 2nd condition is X(a) = B sin (ka) = 0 rightdoublearrow k = n pi/a And the 4th condition implies Y(infinity) = 0 rightdoublearrow C = 0 so, Y(y) = De^-ky And so, the general solution is given by V(x, y) = sigma^infinity_n = 1 A_n sin (n pi x/a) e^-n pi/a y

    $\Rightarrow \frac{\partial^2 V}{\partial x^2}+\frac{\partial^2 V}{\partial y^2}=0$
   We apply the separation of variables
   $V(x,y)=X(x)Y(y)$
And so, the equation reduces to
$\Rightarrow \frac{1}{X}\frac{\mathrm{d}^2 X}{\mathrm{d} x^2}+\frac{1}{Y}\frac{\mathrm{d}^2 Y}{\mathrm{d} y^2}=0$
  
   $\Rightarrow X(x)=A\cos(kx)+B\sin(kx)$
And
   $Y(y)=Ce^{ky}+D^{-ky}$
And given the boundary conditions
   $V(0,y)=0,~~V(a,y)=0,~~~V(x,0)=V_0,~~~V(x,\infty)=0$
And so, the first condition is
   $X(0)=A=0$
And so,
    $\Rightarrow X(x)=B\sin(kx)$
And the 2nd condition is
    $X(a)=B\sin(ka)=0\Rightarrow k=\frac{n\pi}{a}$
And the 4th condition implies
   $Y(\infty)=0\Rightarrow C=0$
So,
   $Y(y)=De^{-ky}$
And so, the general solution is given by
   $\Rightarrow V(x,y)=\sum_{n=1}^{\infty}A_n\sin\left ( \frac{n\pi x}{a} \right )e^{-\frac{n\pi }{a}y}$
And the 3rd condition implies
   $V(x,0)=\sum_{n=1}^{\infty}A_n\sin\left ( \frac{n\pi x}{a} \right )=V_0$
Now we use the Fourier trick to calculate the coefficient A_n. So, we get
    $\sum_{n=1}^{\infty}A_n\int_0^a \sin\left ( \frac{n\pi x}{a} \right )\sin\left ( \frac{m\pi x}{a} \right )dx=V_0\int_0^a \sin\left ( \frac{m\pi x}{a} \right )~dx$
   $\Rightarrow \sum_{n=1}^{\infty}A_n\frac{a}{2}\delta_{m,n}=V_0\frac{a}{m\pi}(1-(-1)^m)$
   $\Rightarrow A_m=\frac{2V_0}{m\pi}(1-(-1)^m)$
      $\Rightarrow A_m=\left\{\begin{matrix} \frac{4V_0}{m\pi}, & m~~odd \\ 0, & m~~even \end{matrix}\right.$
And so,
$\Rightarrow V(x,y)=\sum_{n=1,~odd}^{\infty}\frac{4V_0}{n\pi}\sin\left ( \frac{n\pi x}{a} \right )e^{-\frac{n\pi }{a}y}$