Potential(x,y) Using Laplace's Equation Div(epsilon*gradV)=0

Assume we have a rectangular distribution of dielectric constant, epsilon, inside a rectangle and the potential at the top is V=Vb and the potential at the bottom is V=0. When the charge density, rho, is zero, the differential equation for the potential, V, is Div(epsilon*grad(V))=0. This is the same as saying that epsilon*grad(V)=C where C is a constant. The integral[dV/dy {dy,0,b}]=Vb. In order to efficiently solve for V(x,y) we need to have a good estimate of V(y) for the internal nodes. For our simple case, if there are rows increments in the y direction we might first use the guess Vg: Vg(row)=(Vb/rows)*row; The guess can be greatly improved by making use of the exact 1D solution described below: Since in 1D the Laplace equation is d/dx(epsilon*dV/dx)=0 we know that epsilon*dV/dx)=C, a constant. Since the integral{grad(V) {x,0,xEnd})=Vend, which is known,we can write the following equation for C: C*integral(1/epsilon(x) {x,0,xEnd})=Vend. Then C=Vend/[integral(1/epsilon(x) {x,0,xEnd})] Since we now know C we can write the following equation for 1D V(x): V(x)=C*integral(1/epsilon(x'),{x',0,x}). Therefore we can refine the 2D guess by using the following equation Vg(col,row)=Vb*sum[1/eps(col,row'),{row',0,row}]/sum[1/eps(col,row') {row',0,rows}] which is the same as the 1D solution for V(row); For the finite difference method that I used to refine the guess, I very strongly recommend the following link: https://atmos.washington.edu/~breth/classes/AM585/lect/rjl_585. Since we need the divergence of epsilon*gradient(V) and epsilon(x,y) has x and y variation we have to take the first differentials of epsilon and V in addition to the second differential of V: i.e. d/dx(eps*dV/dx)=eps*d2Vdx2+depx/dx*dV/dx and similarly for the y differential. It is important to notice the way the dielectric slab greatly excludes the electric field, -grad(V), which is inversely proportional the the spacing between the potential contours in the slab. This is similar to the way a superconductor excludes an external magnetic field. In the animation, the slab is oscillated vertically through the electric field in order to see how the potential contours enter and exit the slab. Since the calculation requires several minutes, the frames are saved to an array which can be played back at any time. To avoid confusion, it is better to press the "EtOp" button before playing back the movie.