Capacitor with a Dielectric — Question 3

Question 3

Dielectric layers across the gap divide voltage inversely with permittivity

Two dielectric layers each occupy half the gap d of a plate capacitor of area A. Their relative permittivities are 2 and 6. There is no free charge at their interface; thus $\epsilon_1E_1=\epsilon_2E_2$. Fringing is negligible.

1. Find C in terms of $C_0=\epsilon_0A/d$.

2. At applied voltage V, find both fields and layer voltage drops.

Question 3 - Solution

Dielectric layers across the gap divide voltage inversely with permittivity

(a) Method and calculation

$$V=\frac QA\left(\frac{d/2}{2\epsilon_0}+\frac{d/2}{6\epsilon_0}\right)=\frac{Qd}{3\epsilon_0A}.$$$$C=3C_0.$$

(b) Interpretation and further calculation

$$E_1=\frac{Q}{2\epsilon_0A}=\frac{3V}{2d},\quad E_2=\frac{Q}{6\epsilon_0A}=\frac{V}{2d}.$$$$V_1=3V/4,\quad V_2=V/4.$$

Physical check

The lower-permittivity layer sustains the larger electric field and voltage drop.