Parallel plate capacitor is constructed using three different dielectric materials as shown in fig.

The parallel plates, across which a potential difference is applied are of area A cm² and separated by a distance d cm. The capacitance across AB is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\frac{ε_{0} A}{d} [\frac{2}{K_{1}} + \frac{K_{2} K_{3}}{K_{1} + K_{3}}]$

$\frac{ε_{0} A}{d} [\frac{K_{1}}{2} + \frac{(K_{2} + K_{3})}{K_{2} K_{3}}]$

$\frac{ε_{0} A}{d} [\frac{2}{K_{1}} + \frac{(K_{2} + K_{3})}{K_{1} K_{3}}]$

$\frac{ε_{0} A}{d} [\frac{K_{1}}{2} + \frac{K_{2} K_{3}}{K_{2} + K_{3}}]$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{aligned} C_{1} = \frac{K_{1} ε_{0} (\frac{A}{2})}{d} = \frac{K_{2} ε_{0} A}{d} \\ C_{2} = \frac{K_{2} ε_{0} (\frac{A}{2})}{(\frac{d}{2})} = \frac{K_{2} ε_{0} A}{d} \end{aligned}$
and $C_{3} = \frac{K_{3} ε_{0} A}{d}$
Now $C = C_{1} + \frac{C_{2} C_{3}}{C_{2} + C_{3}}$