Explanation

(b) Equivalent capacitance between points $B\ and\ C$ is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C^{'} = \ \frac{10 \times 10}{10 + 10} + 10 = 15\mu F$
Now equivalent capacitance between points $A\ and\ C$ is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C^{''} = \frac{5 \times 15}{15 + 5} = \frac{75}{20}\mu F$
Charge on capacitor of capacity 5$\mu F$ is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Q = CV = \frac{75}{20} \times 2000 = 7500\mu C$
(Since, potential at the point $C$ will be zero)
Now, potential difference across capacitor of 5$\mu F$ is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ V_{A} - V_{B}\ = \frac{Q}{5\mu F} = \frac{7500\mu C}{5\mu C}$=1500volt
As,$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V_{A} = 2000$volt
Hence, $\ V_{B} = 2000 - 1500 = 500$ volt.