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)

$9C/4$

Answer 2

Area of the given metallic plate

A = πr²

Area of the dielectric plate

$A' = \pi\left( \frac{r}{2} \right)^{2} = \frac{A}{4}$

Uncovered area of the metallic plates

$A" = A - A'$

$= A - \frac{A}{4} = \frac{3A}{4}$

The given situation is equivalent to a parallel combination of two capacitor. One capacitor (C') is filled with a dielectric medium (K = 6) having area $\frac{A}{4}$ while the other capacitor (C'') is air filled having area $\frac{3A}{4}$

Hence $C_{eq} = C' + C" = \frac{K\varepsilon_{0}(A/4)}{d} + \frac{\varepsilon_{0}(3A/4)}{d}$

$= \frac{\varepsilon_{0}A}{d}\left( \frac{K}{4} + \frac{3}{4} \right) = \frac{\varepsilon_{0}A}{d}\left( \frac{6}{4} + \frac{3}{4} \right) = \frac{9}{4}C$ $\left( \because C = \frac{\varepsilon_{0}A}{d} \right)$