Question

Two identical parallel plate capacitors are placed in series and connected to a constant voltage source of $V_0$ volt. If one of the capacitors is completely immersed in a liquid with dielectric $K$, the potential difference between the plates of the other capacitor will change to $\frac{K}{K+n}{V}_0$, then find the value of $n$.

Answer 1

When a capacitor is immersed in a dielectric its new capacitance is $KC$
Hence
$C_{eq}=\frac{C\times KC}{C+KC}$ = $\frac{KC}{(K+1)}$
In series charge $Q$ the capacitors will be same as that on $C_{eq}$
$Q=C_{eq}\times V_0$
Now the potential difference on capcitor without dielectric is
$=\frac{Q}{C}=\frac{C_{eq}{V}_0}{C}=\frac{K{V}_0}{(K+1)}$
$\Rightarrow n=1.00$