Q. Four plates of the same area of cross-section are joined as shown in the figure. The distance between each plate is $d.$ The equivalent capacity across $A$ and $B$ will be
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
(b) The given arrangement is equivalent to the parallel combination of three identical capacitors. Hence equivalent capacitance $= 3C = 3\frac{\varepsilon_{0}A}{d}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.