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 \[C=\frac{{{\varepsilon }_{0}}A}{d}\] \[\frac{{{\varepsilon }_{0}}A}{d}=1\mu F\]                           ?(i) When it is filled with two 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 become half) \[C=\frac{4{{\varepsilon }_{0}}A}{2d}+\frac{6{{\varepsilon }_{0}}A}{2d}\] \[=\frac{2{{\varepsilon }_{0}}A}{d}+3\frac{{{\varepsilon }_{0}}A}{d}\]                 Using Eq. (i), we obtain                 \[C=2\times 1+3\times 1=5\mu F.\]