The capacity of parallel plate condenser is $5 \mu \mathrm{F}$. When a glass plate is placed between the…
The capacity of parallel plate condenser is $5 \mu \mathrm{F}$. When a glass plate is placed between the plates of the condenser, its potential difference reduces to $1 / 8$ of the original value. The magnitude of relative dielectric constant of glass is
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Solution
Given, $C=5 \mu \mathrm{F}=5 \times 10^{-6} \mathrm{~F}$
Initial potential $=V$
When glass plate of dielectric constant $K$ is introduced between the plates of capacitor, then potential difference is
$
\begin{aligned}
& V^{\prime} & =\frac{V}{8} \\
\therefore & K & =\frac{V}{V^{\prime}}=\frac{V}{V / 8} \\
\Rightarrow & K & =8
\end{aligned}
$