Q.
In the figure a capacitor is filled with dielectrics. The resultant capacitance is
A) $\frac{2\varepsilon_{0}A}{d}\left\lbrack \frac{1}{k_{1}} + \frac{1}{k_{2}} + \frac{1}{k_{3}} \right\rbrack$
B) $\frac{\varepsilon_{0}A}{d}\left\lbrack \frac{1}{k_{1}} + \frac{1}{k_{2}} + \frac{1}{k_{3}} \right\rbrack$
C) $\frac{2\varepsilon_{0}A}{d}\left\lbrack k_{1} + k_{2} + k_{3} \right\rbrack$
D) None of these
Explanation
(d) $C_{1} = \frac{K_{1}\varepsilon_{0}\frac{A}{2}}{\left( \frac{d}{2} \right)} = \frac{K_{1}\varepsilon_{0}A}{d}$
$C_{2} = \frac{K_{2}\varepsilon_{0}\left( \frac{A}{2} \right)}{\left( \frac{d}{2} \right)} = \frac{K_{2}\varepsilon_{0}A}{d}$ and $C_{3} = \frac{K_{3}\varepsilon_{0}A}{2d} = \frac{K_{3}\varepsilon_{0}A}{2d}$
Now, $C_{eq} = C_{3} + \frac{C_{1}C_{2}}{C_{1} + C_{2}} = \left( \frac{K_{3}}{2} + \frac{K_{1}K_{2}}{K_{1} + K_{2}} \right).\frac{\varepsilon_{0}A}{d}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.