Q.
A parallel plate capacitor with air as the dielectric has a capacitance $C.$ A slab of dielectric constant $K$ and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure The new capacitance will be
A) $(K + 3)\frac{C}{4}$
B) $(K + 2)\frac{C}{4}$
C) $(K + 1)\frac{C}{4}$
D) $\frac{KC}{4}$
Explanation
(a) The two condensers with $K$ and with air are in parallel
With air, $C_{1} = \frac{\varepsilon_{0}}{d}\left( \frac{3A}{4} \right) = \frac{3\varepsilon_{0}A}{4d}$
With medium, $C_{2} = \frac{\varepsilon_{0}K}{d}\left( \frac{A}{4} \right) = \frac{\varepsilon_{0}AK}{4d}$
$\therefore C^{'} = C_{1} + C_{2}$
$\Rightarrow C^{'} = \frac{3\varepsilon_{0}A}{4d} + \frac{\varepsilon_{0}AK}{4d} = \frac{\varepsilon_{0}A}{4}\left\lbrack \frac{3}{4} + \frac{K}{4} \right\rbrack = \frac{C}{4}(K + 3)$
$\Rightarrow C^{'} = \frac{C}{4}(K + 3)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.