Q. Two identical parallel plate capacitors are placed in series and connected to a constant voltage source of $V$ volt. If one of the capacitor is completely immersed in a liquid of dielectric constant $K$, then the potential difference between the plates of the other capacitor will change to
A) $\frac{K}{K + 1}V$
B) $\frac{K + 1}{K}$
C) $\frac{2K}{K + 1}V$
D) $\frac{K + 1}{2K}V$
Explanation
(a) In series combination, charge is constant.
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V \propto \frac{1}{C}$
Now, $\ \ \ \ \ \ \ \ \ \ \frac{V_{2}}{V_{1}} = \frac{KC}{C} = \frac{K}{1}$
But $\ \ \ \ \ \ \ \ \ \ V_{1} + V_{2}\ = V$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{V_{2}}{K} + V_{2}\ = V\ or\ V_{2} = \frac{K}{K + 1}V\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.