Q.
Two parallel plates of area $A$ are separated by two different dielectric as shown in figure. The net capacitance is
A) $\frac{\varepsilon_{0}A}{2d}$
B) $\frac{\varepsilon_{0}A}{d}$
C) $\frac{3\varepsilon_{0}A}{d}$
D) $\frac{4\varepsilon_{0}A}{3d}$
Explanation
(d) $C = \frac{\varepsilon_{0}A}{\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}}}$
$= \ \frac{\varepsilon_{0}A}{\frac{d}{2}\left( \frac{1}{1} + \frac{1}{2} \right)} = \frac{4\varepsilon_{0}A}{3d}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.