Q. The potential energy of a charged parallel plate capacitor is $U_{0}$. if a slab of dielectric constant $k$ is inserted between the plates, than the new potential energy will be
A) $\frac{U_{0}}{k}$
B) $U_{0}k^{2}$
C) $\frac{U_{0}}{k^{2}}$
D) $U_{0}^{2}$
Explanation
(a) $P.E.U_{o} = \frac{Q^{2}}{2C}$
When a slab of dielectric constant $k$ is inserted,
Then $C^{'} = Ck$
$U^{'} = \frac{Q^{2}}{2C'} = \frac{Q^{2}}{2Ck} = \frac{U_{o}}{k}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.