A cubical box with sides of length L consists of six metal plates. Five sides of the box { the plates at $x=0, x=L, y=0, y=L, z=0$ - are grounded. The top of the box (at z = L) is made of a separate sheet of metal, insulated from the others, and held at a constant potential $V_0$. Find the potential inside the box.

I am attempting to set up this problem with Legendre polynomials. Since the top portion is grounded, this affects the "potential" region I can take, and I am unclear on how about to go setting it up.

I have set up the Laplace as follows:

$1\over x$${d^2x\over dx^2} +$ $1\over y$${d^2y\over dy^2} +$$1\over z$${d^2z\over dz^2} = 0$

In which at this point, must each component equal a constant?

which gives $k_x^2+k_y^2+k_z^2 = 0$

$1\over x$${d^2x\over dx^2} = k_x^2$

$1\over y$${d^2y\over dy^2} = k_y^2$

$1\over z$${d^2z\over dz^2} = k_z^2$

where z is given by: $Z(z) = sinh(k_zz)$

but at this point, how do I solve the coefficients?

Or should I be using the general form and find coefficients:

$x = Acos(k\theta) + Bsin(k\theta)$

$y = Ccos(m\phi) + Dsin(m\phi)$

$z = Ee^{(k^2 +m^2)^.5z}$ + $Fe^{-(k^2 +m^2)^.5z}$

I'm stuck on finding the coefficients using this method.


3 Answers 3


This is a rather simple problem. Such problems are best approached by using the ansatz approach - meaning trying to guess the form the solution will take, and if your guess for the solution satisfies both Laplace's equation and the boundary conditions then the uniqueness theorem guarantees that you have found THE solution. Now, regarding your problem, the zero boundary conditions for x and y are reminiscent of standing waves with nodes at both ends. So we can guess the x and y dependencies to be $sin\frac{m \pi x}{L}$ and $sin\frac{n \pi y}{L}$ respectively. Your statement $k_x^2 + k_y^2 + k_z^2 = 0$ doesn't look right since the sum of three positive real nos. cannot be zero. Rather, you would have $$ -(\frac{m \pi}{L})^2 - (\frac{n \pi}{L})^2 + A_z = 0 $$. Since, the sign of the constant in the equations for '$X$' and '$Y$' are negative, the constant for the '$Z$' differential equation would automatically be positive. This would in turn indicate hyperbolic sine or cosine functions since these functions are the solutions to the differential equation with positive value for the coefficient. The boundary condition of zero at $z=0$ indicates that only the hyperbolic sine function will play a role. The complete solution would thus be given by $$\phi(x,y,z) = \sum_{m,n = 1}^\infty C_{m,n} sin\frac{m \pi x}{L} sin\frac{n \pi y}{L} sinh \frac{\sqrt(m^2+n^2) \pi z}{L}$$. The coefficients $C_{m,n}$ can be determined by applying the boundary condition at $z = L$ and using the orthogonality relations for the sine functions.


You have to solve the Laplace equation with the given boundary conditions. In cartesian coordinates, the solution has the form V = X(x)Y(y)Z(z).


The basic steps:

  1. Write the potential $V(x,y,z)$ as a series in products of appropriate orthogonal functions. Often one uses orthogonal functions that arise from the solution to Laplace's equation by separation of variables.

  2. Impose the boundary conditions (there are six of them) to solve for the coefficients in the series expansion. For example, the $z=L$ boundary condition would be $$ V(x,y,L) = V_0 $$ All "grounded" sides are at zero potential; this is just the mathematical definition of the term "grounded." So, for example, at $z=0$ you would have $$ V(x,y,0) = 0 $$


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