Two identical parallel plate air capacitors are connected in series to a battery of emf ' $\mathrm{V}$ '. If…
Two identical parallel plate air capacitors are connected in series to a battery of emf ' $\mathrm{V}$ '. If one of the capacitor is inserted in liquid of dielectric constant ' $\mathrm{K}$ ' then, potential difference of the other capacitor will become
$\frac{\mathrm{K}-1}{\mathrm{KV}}$
$\frac{\mathrm{K}+1}{\mathrm{KV}}$
$\left(\frac{\mathrm{KV}}{\mathrm{K}+1}\right)$
$\frac{\mathrm{KV}}{\mathrm{K}-1}$
Solution
Let the capacitance of capacitors are C initially.
So, afterward $\mathrm{C}_1=\mathrm{KC}$ and $\mathrm{C}_2=\mathrm{C}$
p.d. across capacitor $\mathrm{C}_2$ is
$\mathrm{V}_2=\frac{\mathrm{C}_1 \mathrm{~V}}{\mathrm{C}_1+\mathrm{C}_2}=\frac{\mathrm{KCV}}{\mathrm{C}+\mathrm{KC}}=\frac{\mathrm{KV}}{\mathrm{K}+1}$