Q. Four plates of equal area $A$ are separated by equal distances $d$ and are arranged as shown in the figure. The equivalent capacity is
A) $\frac{2\varepsilon_{0}A}{d}$
B) $\frac{3\varepsilon_{0}A}{d}$
C) $\frac{4\varepsilon_{0}A}{d}$
D) $\frac{\varepsilon_{0}A}{d}$
Explanation
(a) The given circuit is equivalent to a parallel combination two identical capacitors
Hence equivalent capacitance between $A$ and $B$ is
$C = \frac{\varepsilon_{0}A}{d} + \frac{\varepsilon_{0}A}{d}$
$= \frac{{2\varepsilon}_{0}A}{d}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.