# Capacitor with 2 dielectrics

1. Feb 26, 2008

### iamaelephant

[SOLVED] Capacitor with 2 dielectrics

1. The problem statement, all variables and given/known data
A parallel plate capacitor of plate area A and spacing d is filled with two parallel slabs of dielectric of equal thickness with dielectric constants k1 and k2, respectively. What is the capacitance?

2. Relevant equations
You tell me.

3. The attempt at a solution
I wasn't sure if I could do something simple like take an average of the dielectric constants (although I suspect not). Instead I split the capacitor into two series capacitors and added them in the usual way. Is this allowed? Off the top of my head it seems to be okay, but I'm probably wrong.

In any case, here is my working. This electromagnetic stuff is definitely my weak hand so advice on these types of problems would be great.

Splitting the capacitor into two series capacitors, we can get their capacitances using the equation
$$C_1 = \frac{k_1 \epsilon_0 A}{\frac{d}{2}} = \frac{2k_1 \epsilon_0 A}{d}$$
$$C_2 = \frac{k_2 \epsilon_0 A}{\frac{d}{2}} = \frac{2k_2 \epsilon_0 A}{d}$$

$$\frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{2 k_1 \epsilon_0 A} + \frac{d}{2 k_2 \epsilon_0 A} = \frac{2 d \epsilon_0 A (k_1 + k_2)}{(2 k_1 \epsilon_0 A)(2 k_2 \epsilon_0 A)}$$

$$C_t = \frac{2 \epsilon_0 A k_1 k_2}{d(k_1 + k_2}$$

2. Feb 26, 2008

### G01

This is the correct way to go about the problem. The capacitor filled with two dielectrics can be considered to be two capacitors in series with different dielectrics.

3. Feb 26, 2008

### iamaelephant

Excellent, thanks so much. I'm stoked I got this one right, a rare hit for me in EM :P