Q. A parallel plate capacitor is charged. If the plates are pulled apart
A) The capacitance increases
B) The potential difference increases
C) The total charge increases
D) The charge and potential difference remain the same
Explanation
(b) Capacitance of a parallel plate capacitor $C = \frac{\varepsilon_{0}A}{d}$
$\because$ Potential difference = $\frac{Charge}{Capacitance}$
$\therefore$ Potential difference increases
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.