question_answer

Consider a parallel plate capacitor of \[10\mu F\] (micro-farad) with air filled in the gap between the plates. Now one-half of the space between the plates is filled with a dielectric of dielectric constant 4, as shown in the figure. The capacity of the capacitor changes to

A) \[25\mu F\]                       

B) \[20\mu F\]

C) \[40\mu F\]                       

D) \[5\mu F\]

Correct Answer: A

Solution :

\[{{C}_{1}}=\frac{{{\varepsilon }_{0}}\left( \frac{A}{4} \right)}{d},\,\,{{C}_{2}}=\frac{K{{\varepsilon }_{0}}\left( \frac{A}{2} \right)}{d}\]                 \[{{C}_{3}}=\frac{{{\varepsilon }_{0}}\left( \frac{A}{4} \right)}{d}\]                 \[{{C}_{eq}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}\]                 \[=\left( \frac{K+1}{2} \right)\frac{{{\varepsilon }_{0}}A}{d}\]                 \[=\left( \frac{4+1}{2} \right)\times 10=25\mu F\]