question_answer

A capacitor of capacitance 1 \[\mu F\] is filled with two dielectrics of dielectric constants 4 and 6. What is the new capacitance?

A)  10\[\mu F\]                      

B)         5\[\mu F\]

C)  4\[\mu F\]                        

D)         7\[\mu F\]

Correct Answer: B

Solution :

Initially, the capacitance of capacitor                 \[C=\frac{{{\varepsilon }_{0}}A}{d}\] \[\therefore \]  \[\frac{{{\varepsilon }_{0}}A}{d}=1\mu F\]                                           ... (i) When it is filled with dielectrics of dielectric constants\[{{K}_{1}}\]and\[{{K}_{2}}\]as shown, then there are two capacitors connected in parallel. So,                 \[C=\frac{{{K}_{1}}{{\varepsilon }_{0}}(A/2)}{d}+\frac{{{K}_{2}}{{\varepsilon }_{0}}(A/2)}{d}\]                                               (as area becomes half)                 \[C=\frac{4{{\varepsilon }_{0}}A}{2d}+\frac{6{{\varepsilon }_{0}}A}{2d}\]                      \[=\frac{2{{\varepsilon }_{0}}A}{d}+\frac{3{{\varepsilon }_{0}}A}{d}\] Using Eq. (i), we obtain                 \[C=2\times 1+3\times 1=5\mu F\]