Question

The two metallic plates of radius $r$ are placed at a distance $d$ apart and its capacity is $C$. If a plate of radius $r/2$ and thickness $d$ of dielectric constant 6 is placed between the plates of the condenser, then its capacity will be

• A. 7C/2
• B. 3C/7
• C. 7C/3
• D. 9C/4

Answer 1

Answer: (D)

Area of the given metallic plate, $A=\pi r^2$
Area of the dielectric plate, $A^{\prime }=\pi {\left(\frac{r}{2}\right)}^2=\frac{A}{4}$
Uncovered area of the metallic plates, $A^{\prime \prime }=A-A^{\prime }$
$=A-\frac{A}{4}=\frac{3A}{4}$

The given situation is equivalent to a parallel combination of two capacitor. One capacitor $\left(C^{\prime }\right)$ is filled with a dielectric medium $(K=6)$ having area $\frac{A}{4}$ while the other capacitor $\left(C^{\prime }\right)$ is air filled having area $\frac{3A}{4}$.
Hence, $C_{eq}=C^{\prime }+C^{\prime \prime }=\frac{K{\epsilon }_0(A/4)}{d}+\frac{{\epsilon }_0(3A/4)}{d}$
$=\frac{{\epsilon }_{0}A}{d}\left(\frac{K}{4}+\frac{3}{4}\right)$
$=\frac{{\epsilon }_{0}A}{d}\left(\frac{6}{4}+\frac{3}{4}\right)$
$=\frac{9}{4}C\left(\because C=\frac{{\epsilon }_{0}A}{d}\right)$