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
Text Solution
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Area of the given metallic plate A = π π r 2
Area of the dielectric plate $A' = \pi \left(\frac{r}{2}\right)^2 = \frac{A}{4}$
Uncovered area of the metallic plates \vec{A}'' = \vec{A} - \vec{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 $C_{eq} = C' + C'' = \frac{K \varepsilon_0 (A/4)}{d} + \frac{\varepsilon_0 (3A/4)}{d}$
$= \frac{\varepsilon_0 A}{d} \left( \frac{K}{4} + \frac{3}{4} \right)$ $\frac{\epsilon_0 A}{d} \left( \frac{6}{4} + \frac{3}{4} \right) = \frac{9}{4} C$ $\left( \boxed{\times} C = \frac{\varepsilon_0 A}{d} \right)$
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