In the circuit shown, a potential difference of 60V is applied across AB. The potential difference between the points M and N is 

The capacitors 2C, C and 2C are in series. Their equivalent is given by
$\frac{1}{C_0}=\frac{1}{2C}+\frac{1}{C}+\frac{1}{2C}$
Or,  $C_0=\frac{C}{2}$
∴ charge on each of these three capacitors is 
$q=C_{0}V=\frac{C}{2}\left(60\right)=30C$.
∴ potential difference across M and N = potential difference across the capacitance $C=\frac{q}{C}=30V$ .