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

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