Q.
Four plates of equal area $A$ are separated by equal distance $d$ 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) $\frac{3\varepsilon_{0}A}{2d}$
D) $\frac{\varepsilon_{0}A}{d}$
Explanation
(a) The given circuit is equivalent to a parallel combination of two identical capacitors.
[image:2_image101.emf]
Hence, equivalent capacitance between points $A\ and\ B$ 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.