question_answer

A capacitor having capacitance \[1\,\mu F\]with air, is filled with two dielectrics as shown. How many times   capacitance will increase?

A)  12

B)  6

C)  8/3

D)  3

Correct Answer: B

Solution :

 Key Idea: After filling with dielectrics the two capacitors will be in parallel order. As shown, the two capacitors are connected in parallel. Initially the capacitance of capacitor \[C=\frac{{{\varepsilon }_{0}}A}{d}\] where A is area of each plate and d is the separation between the plates. After filling with dielectrics, we have two capacitors of capacitance. \[{{C}_{1}}=\frac{{{K}_{1}}{{\varepsilon }_{0}}(A/2)}{d}=\frac{8}{2}\frac{{{\varepsilon }_{0}}A}{d}=\frac{4{{\varepsilon }_{0}}A}{d}=4C\] and \[{{C}_{2}}=\frac{{{K}_{2}}{{\varepsilon }_{0}}(A/2)}{d}=\frac{4}{2}\frac{{{\varepsilon }_{0}}A}{d}=\frac{2{{\varepsilon }_{0}}A}{d}=2C\] Hence, their equivalent capacitance \[{{C}_{eq}}={{C}_{1}}+{{C}_{2}}\] \[=4C+2C=6C\] i.e., new capacitance will be six times of the original.