Q. Two condensers of capacities $2C$ and $C$ are joined in parallel and charged upto potential $V.$ The battery is removed and the condenser of capacity $C$ is filled completely with a medium of dielectric constant $K.$ The p.d. across the capacitors will now be
A) $\frac{3V}{K + 2}$
B) $\frac{3V}{K}$
C) $\frac{V}{K + 2}$
D) $\frac{V}{K}$
Explanation
(a) $q_{1} = 2CV,\ q_{2} = CV$
Now condenser of capacity $C$ is filled with dielectric $K,$
therefore $C_{2} = KC$
As charge is conserved
$\therefore q_{1} + q_{2} = \left( C_{2}^{'} + 2C \right)V^{'} \Rightarrow V^{'} = \frac{3CV}{(K + 2)C} = \frac{3V}{K + 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.