Two identical parallel plate air capacitors are connected in series to a battery of e..m.f. ${ }^{\prime}…
Two identical parallel plate air capacitors are connected in series to a battery of e..m.f. ${ }^{\prime} \mathrm{V}^{\prime}$. If one of the capacitor is inserted in liquid of dielectric constant '$\mathrm{K}^{\prime}$, then potential difference of the other capacitor will become
$\frac{\mathrm{K}}{\mathrm{V}(\mathrm{K}+1)}$
$\frac{\mathrm{KV}}{\mathrm{K}+1}$
$\frac{\mathrm{K}+1}{\mathrm{KV}}$
$\frac{\mathrm{K}}{\mathrm{V}(1-\mathrm{K})}$
Solution
Let the capacitance of capacitor \(C_{2}\) be \(C\). Thus capacitance of capacitor \(C_{1}\) is KC.
Potential across the capacitor \(\mathrm{C}_{2}, \mathrm{~V}_{2}=\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}} \mathrm{~V}\)
\(\therefore \mathrm{V}_{2}=\frac{\mathrm{KC}}{\mathrm{C}+\mathrm{KC}} \mathrm{V}\)
\(\Longrightarrow \mathrm{V}_{2}=\frac{\mathrm{KV}}{\mathrm{K}+1}\)
.