Four capacitors with capacitances $C_1=1\mu F,$$C_2=1.5\mu F,$$C_3=2.5\mu F$$\mathrm{and}$$C_4=0.5\mu F$  are connected as shown and are connected to 30 V source. The potential difference between a and b is
 

Let $\text{ q }\mu \text{C}$  be the change on two capacitors  

Then    

$\text{⇒q=18 μC}$

$V_A-V_a=\frac{q}{C_1}=\frac{18}{1}=18 \mathrm{V}$

$\text{ Similarly, }\frac{q^{\text{'}}}{C_3}+\frac{q^{\text{'}}}{C_4}=30\text{ or }\frac{q^{\text{'}}}{2.5}+\frac{q^{\text{'}}}{0.5}=30$

$q\text{'}=\frac{25}{2}\mu C$

$V_A-V_b=\frac{q^{\text{'}}}{C_3}=\frac{25}{2\times 2.5}=5V$

$V_a-V_b=\left(V_A-V_b\right)-\left(V_A-V_a\right)=5-18=-13 V$