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 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 becomes half)                 \[C=\frac{4{{\varepsilon }_{0}}A}{2\,d}=3\frac{6{{\varepsilon }_{0}}A}{2\,d}\]                 \[=\frac{2{{\varepsilon }_{0}}A}{d}+3\frac{{{\varepsilon }_{0}}A}{d}\] Using Eq. (i), we obtain \[C=2\times 1+3\times 1=5\,\mu F\].