Question

A capacitor has capacitance \(C_0\) when there is no dielectric between its plates. Two slabs of dielectric constant \(K_1\), \(K_2\) respectively with area equal to area of plates but thickness half of the distance between the plates are placed in between the plates. Then the new capacitance is

• A. \(C_0 \left( \frac{K_1 + K_2}{2} \right)\)
• B. \(C_0 \left( \frac{2 K_1 K_2}{K_1 + K_2} \right)\)
• C. \(C_0 \left( \frac{K_1 K_2}{K_1 + K_2} \right)\)
• D. \(2 C_0 \left( \frac{K_1 K_2}{K_1 + K_2} \right)\)

Hint:

Treat dielectric slabs as capacitors in series and apply the formula \(\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}\).

Answer 1

The Correct Option is D

The capacitor is effectively two capacitors in series, each with dielectric \(K_1\) and \(K_2\) and thickness \(d/2\). Using the formula for capacitors in series: \[ \frac{1}{C} = \frac{d}{2 \epsilon_0 K_1 A} + \frac{d}{2 \epsilon_0 K_2 A} = \frac{d}{2 \epsilon_0 A} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) \Rightarrow C = \frac{2 \epsilon_0 A}{d} \left( \frac{K_1 K_2}{K_1 + K_2} \right) = 2 C_0 \left( \frac{K_1 K_2}{K_1 + K_2} \right) \]