# Soln of Leplace in Cartesian Coord

## Main Question or Discussion Point

Hi

I'm getting confused solving the Laplace eqn in Cartesian coordinates.

The equation can be solved by solving each of
$\frac{X''(x)}{X(x)}=-k_x^2, \qquad\qquad \frac{Y''(y)}{Y(y)}=-k_y^2, \qquad\qquad \frac{Z''(z)}{Z(z)}=k_z^2$

and then substituting into the equation
$\Phi(\vec{x})=X(x)Y(y)Z(z)$

The solution to the (general) ODE's is
$X(x)=A_1e^{ik_xx}+A_2e^{-ik_xx}, \qquad\qquad Y(y)=B_1e^{ik_yy}+B_2e^{-ik_yy}, \qquad\qquad Z(z)=C_1e^{k_zz}+C_2e^{k_zz}$

But the solution when solving for the electric potential is given by
$\Phi_{k_x,k_y,\pm}(\vec{x})=exp\left(ik_xx+ik_yy \pm\sqrt{k_x^2+k_y^2}\right)$

Where am I going wrong?
Why have they let the coefficients $A_{1,2}, B_{1,2}, C_{1,2}=1$?

Thanks

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The values for the coefficients in the general solution can be found by applying the boundary conditions of the problem. Nobody can answer your question without more information about what the boundary conditions were. (You need to give us the whole problem.) Depending on the geometry of the problem, certain conditions will force certain coefficients to have particular values, which will usually simplify your solution to something with a manageable form. You will normally end up doing something like Fourier's trick to determine the value of the coefficient determined by the last boundary condition, and you will end up with an expression for the potential. Sometimes, things will simplify really nicely.

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So then is the general solution always

$\Phi(\vec{x})=\left[A_1e^{ik_xx}+A_2e^{ik_xx}\right]\cdot\left[B_1e^{ik_yy}+B_2e^{ik_yy}\right]\cdot\left[C_1e^{k_zz}+C_2e^{-k_zz}\right]$

and depending on the boundary condition we solve for the $A_{1,2}, B_{1,2}, C_{1,2}, k_1 \quad\mbox{and}\quad k_2$?

The problem I have is solving the electric potential of a uniform field, when a dielectric cube is placed within it. I can assume that the field lines are parallel to two of the cube's sides.

If you know how I can approach this problem, I would be grateful if you could guide me.