Question

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 $\mu F$

Answer 1

Answer: (B)

Initially, the capacitance of capacitor
$C=\frac{{\epsilon }_{0}A}{d}$
$\therefore$ $\frac{{\epsilon }_{0}A}{d}=1\mu F$ ... (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, $C=\frac{K_1{\epsilon }_0(A/2)}{d}+\frac{K_2{\epsilon }_0(A/2)}{d}$ (as area becomes half)
$C=\frac{4{\epsilon }_{0}A}{2d}+\frac{6{\epsilon }_{0}A}{2d}$
$=\frac{2{\epsilon }_{0}A}{d}+\frac{3{\epsilon }_{0}A}{d}$
Using Eq. (i),
we obtain $C=2\times 1+3\times 1=5\mu F$