Q.
Three plates of common surface area $A$ are connected as shown. The effective capacitance will be
A) $\frac{\varepsilon_{0}A}{d}$
B) $\frac{3\varepsilon_{0}A}{d}$
C) $\frac{3}{2}\frac{\varepsilon_{0}A}{d}$
D) $\frac{2\varepsilon_{0}A}{d}$
Explanation
(d) Here, we have two capacitors I and II connected in parallel order.
[image:2_image117.emf]
So, $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C = C_{1} + C_{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{\varepsilon_{0}A}{d} + \frac{\varepsilon_{0}A}{d} = \frac{{2\varepsilon}_{0}A}{d}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.