Q.
A parallel plate capacitor with air as the dielectric has 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
Explanation
(a) The capacitor with air as the dielectric has capacitance
$\ \ \ \ \ \ \ \ \ \ \ \ \ C_{1} = \frac{\varepsilon_{0}}{d}\left( \frac{3A}{4} \right) = \frac{3\varepsilon_{0}A}{4d}$
Similarly, the capacitor with $K$ as the dielectric constant has capacitance $\ \ \ \ \ \ \ \ \ \ \ \ C_{2} = \frac{\varepsilon_{0}K}{d}\left( \frac{A}{4} \right) = \frac{\varepsilon_{0}AK}{4d}$
Since, $C_{1}and\ C_{2}$ are in parallel
$\ \ \ \ \ \ \ \ \ \ \ C_{net} = C_{1} + C_{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{3\varepsilon_{0}A}{4d} + \frac{\varepsilon_{0}AK}{4d}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{\varepsilon_{0}A}{d}\left\lbrack \frac{3}{4} + \frac{K}{4} \right\rbrack\ $
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{C}{4}(K + 3)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.