Q. The capacitance of a parallel plate capacitor is $12\mu F.$ If the distance between the plates is doubled and area is halved, then new capacitance will be
A) $8\ \mu F$
B) $6\ \mu F$
C) $4\ \mu F$
D) $3\ \mu F$
Explanation
(d) $C = \frac{\varepsilon_{0}A}{d}$ As $A \rightarrow \frac{1}{2}$ times and $d \rightarrow 2$ times
So $C \rightarrow \frac{1}{4}$ times $i.e.\ C^{'} = \frac{1}{4}C = \frac{12}{4} = 3\ \mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.