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 [AFMC 2001; MP PET 2001]

A)            \[25\,\mu \,F\]

B)            \[20\,\mu \,F\]

C)            \[40\,\mu \,F\]

D)            \[5\,\mu \,F\]

Correct Answer: A

Solution :

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