The two metallic plates of radius r are place | vedclass

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Question

(d) Area of the given metallic plate $A = \pi r^2$
Area of the dielectric plate $A' = \pi \,{\left( {\frac{r}{2}} \right)^2} = \frac{A}{4}$
Unco

Vedclass · Accepted Answer

$9C/4$

Vedclass · 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 - \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
$ = \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$

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

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