The distance between the plates of a condenser is reduced to $\frac{1}{4} t h$ and the space between the plates is filled up by a medium of dielectric constant K(2.8). The capacity is increased by :

5.6times

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11.2times

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22.4 times

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44.8 times

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Solution

The correct option is B 11.2times
Initially,
$C = \frac{ϵ_{0} A}{d}$
$\Rightarrow$ Distance is reduced by $\frac{1}{4}$ and K is added
$∴ C_{n e w} = \frac{k ϵ_{0} A}{(^{d} /_{a})}$
$\Rightarrow C_{n e w} = 2.8 \times \frac{ϵ_{0} A}{d} \times 4$
$\Rightarrow C_{n e w} = 11.2 \times C .$

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The distance between the plates of a condenser is reduced to 14th and the space between the plates is filled up by a medium of dielectric constant K(2.8). The capacity is increased by :

The correct option is B 11.2timesInitially, C=ϵ0 Ad⇒Distance is reduced by 14 and K is added∴Cnew=k ϵ0 A(d/a)⇒Cnew=2.8×ϵ0 Ad×4⇒Cnew=11.2×C.

The distance between the plates of a condenser is reduced to $\frac{1}{4} t h$ and the space between the plates is filled up by a medium of dielectric constant K(2.8). The capacity is increased by :

The correct option is B 11.2times
Initially,
$C = \frac{ϵ_{0} A}{d}$
$\Rightarrow$ Distance is reduced by $\frac{1}{4}$ and K is added
$∴ C_{n e w} = \frac{k ϵ_{0} A}{(^{d} /_{a})}$
$\Rightarrow C_{n e w} = 2.8 \times \frac{ϵ_{0} A}{d} \times 4$
$\Rightarrow C_{n e w} = 11.2 \times C .$