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) The electric field between the plates is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \frac{V}{d}$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V = \ Ed\ or\ V \propto d\ \ $
Hence, if the plates are pulled apart the potential difference increases.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.