A capacitor of capacitance C is charged to a potential difference V0. The charging battery is disconnected and the capacitor is connected to a capacitor of unknown capacitance Cx. The P.D. across the combination is V. The value of Cx should be

A capacitor of capacitance C having potential difference $V_o$ has charge $Q=C{V}_o Joules$. When the battery is disconnected and unknown capacitor of capacitance $C_x$ is connected charge will be distributed in such a way t​hat it remains conserved.The measued potential difference of both capacitor is V and has net charge Q.So net capacitance is 

$C_{net}=\frac{Q}{V}=\frac{C{V}_o}{V}$

And we also know the series and parallel combination for capacitors so

(i)If connected in series $\frac{1}{C_{net}}=\frac{1}{C}+\frac{1}{C_x}$ Using previous values and substituting

$$\begin{gathered} \frac{V}{C{V}_o}=\frac{1}{C}+\frac{1}{C_x} \\ Multiplying whole equation by C{C}_{x } we get \\ \frac{V{C}_x}{V_o}=C_x+C \\ {C}_x\left(\frac{V}{V_o}-1\right)=C \\ {C}_x=\frac{C{V}_o}{V-V_o} farad \end{gathered}$$

(ii)IF connected in parallel
 
$$\begin{gathered} C_{net}=C+C_x Substituting the previous values we have \\ \frac{C{V}_o}{V}=C+C_x \\ {C}_x=C\left(\frac{V_o}{V}-1\right)=C\left(\frac{V_o-V}{V}\right) farad \end{gathered}$$