Question

A parallel plate capacitor of capacitance $C$ (without dielectrics) is filled by dielectric slabs as shown in the figure. Then the new capacitance of the capacitor is:

• A. 3.9 $C$
• B. 4 $C$
• C. 2.4 $C$
• D. 3 $C$

Answer 1

Answer: (B)

4 $C$

Answer 2

The capacitor can be considered as two capacitors in series. The top capacitor has a dielectric constant of 6 and thickness d. The bottom capacitor is made of two capacitors in parallel, with dielectric constants 2 and 4, each with thickness d.

The capacitance of the top capacitor is $C_{top} = \frac{6\epsilon_0 A}{d}$. The capacitance of the left part of the bottom capacitor is $C_1 = \frac{2\epsilon_0 (A/2)}{d} = \frac{\epsilon_0 A}{d}$. The capacitance of the right part of the bottom capacitor is $C_2 = \frac{4\epsilon_0 (A/2)}{d} = \frac{2\epsilon_0 A}{d}$. The capacitance of the bottom capacitor is $C_{bottom} = C_1 + C_2 = \frac{\epsilon_0 A}{d} + \frac{2\epsilon_0 A}{d} = \frac{3\epsilon_0 A}{d}$.

The equivalent capacitance is given by $\frac{1}{C_{eq}} = \frac{1}{C_{top}} + \frac{1}{C_{bottom}} = \frac{d}{6\epsilon_0 A} + \frac{d}{3\epsilon_0 A} = \frac{d}{6\epsilon_0 A} + \frac{2d}{6\epsilon_0 A} = \frac{3d}{6\epsilon_0 A} = \frac{d}{2\epsilon_0 A}$. Therefore, $C_{eq} = \frac{2\epsilon_0 A}{d}$.

The original capacitance is $C = \frac{\epsilon_0 A}{2d}$. Thus, $C_{eq} = \frac{2\epsilon_0 A}{d} = 4\frac{\epsilon_0 A}{2d} = 4C$.