# Homework Help: Capacitance of a parallel plate capacitor with 2 dielectrics

1. Feb 1, 2010

### bigevil

1. The problem statement, all variables and given/known data

Find the capacitance of the parallel plate capacitor with 2 dielectrics below. Given that the parallel plate capacitor has area A = WL and the separation between plates is d.

3. The attempt at a solution

My method is to use integration. First solve for two small capacitors dC1, dC2 and then sum them up in parallel.

$$y = \frac{d}{L}x$$

$$dC_1 = \frac{\kappa_1 \epsilon_0 WL dx}{d(L-x)}$$

$$dC_2 = \frac{\kappa_2 \epsilon_0 WL}{d} \frac{1}{x} dx$$

$$dC = \frac{dC_1 dC_2}{dC_1 + dC_2} = \frac{\kappa_1 \kappa_2 \epsilon_0 WL}{d} \cdot \frac{1}{\kappa_2 L + x(\kappa_1 - \kappa_2)}$$

$$C = \frac{\kappa_1\kappa_2 \epsilon_0 WL}{d(\kappa_1 - \kappa_2)} \left[ ln (\kappa_2 L + x(\kappa_1 - \kappa_2) \right]_0^L$$

The expression is definitely wrong. If we let the two dielectrics have equal value k, we should expect C to follow that of a single dielectric, C' = kC.

2. Feb 2, 2010

### ehild

You have one more L in the expressions of dC. The area of an elementary capacitor is Wdx.

ehild

3. Feb 2, 2010

### bigevil

Ok, but L is a constant and having amended that, it doesn't make the final integral any more correct. The ln form of the last equation is still clearly wrong.

4. Feb 2, 2010

### ehild

Are you sure that the ln is wrong? You can not simply substitute K1=K2, it will lead to 0/0. You have to take the limit K2/K1-->1.

ehild

Last edited: Feb 2, 2010