Capacitor with multiple dielectrics [closed]

Question

Let's say there is a capacitor as shown in the figure, $K_1$, $K_2$, $K_3$, $K_4$ being the dielectric constant of each quadrant. (the dimensions of all four parts are equal.)

Let $C_1$, $C_2$, $C_3$, $C_4$ be the capacitance of the respective dielectrics, $A$ be the area of cross section of all four dielectrics and $d$ be the width of all the four capacitors. I am supposed to find the equivalent capacitance of this combination.

I was able to come up with two ways to solve this question:

1. Take $K_1$, $K_2$ in series and $K_3$, $K_4$ in series, add the two equivalent dielectrics in parallel.

2. Take $K_1$, $K_3$ in parallel and $K_2$, $K_4$ in parallel, add the two equivalent dielectrics in series.

Should I solve it by method 1 or 2, or are both correct, or is there some other method that I am not aware of?

Answer 1

Both approaches replace the original configuration with a supposedly-equivalent circuit network of four capacitors, but they differ in the way the capacitors are interconnected.

In an electrostatic field, we can introduce a conductor along surfaces where there is the same potential, and the electrostatic field does not change at all. In a parallel-plates capacitor (as usual, ignoring the field distortion that happens at the plate borders) that means that at any distance, we can introduce a separating pseudo-plate and treat the capacitor as a series-combination of the two parts.

On the other hand, we can cut the surface area of a capacitor and treat it as a parallel-combination of the parts.

Now we have to combine these two operations for our problem.

Your first approach first cuts the surface in two parts (one being K1 and K2, and the other K3 and K4), and then introduces two separating plates, one between K1 and K2, and one between K3 and K4. That's perfectly valid.

The second approach differs in the one fact that now you connect the two separating plates, assuming that the have the same potential, which generally isn't true, only in the special case that K1/K2 = K3/K4.

Take K1, K2 in series and K3,K4 in series, add the two equivalent dielectrics in parallel.

This is the correct approach.