For the circuit shown in the related figure, if a capacitor having...

Let the potential at point B be $V_B = 0\text{ V}$.

In steady state, the capacitor acts as an open circuit, so it does not draw any DC current. The current distribution in the resistor network remains unchanged.

Applying Kirchhoff's Current Law (KCL) at node A:

$\dfrac{V_A - 5}{10} + \dfrac{V_A - 5}{10} + \dfrac{V_A - 0}{20} = 0$

Multiplying the equation by $20$, we get:

$2(V_A - 5) + 2(V_A - 5) + V_A = 0$

$4V_A - 20 + V_A = 0$

$5V_A = 20 \Rightarrow V_A = 4\text{ V}$

The potential difference across the capacitor connected between A and B is $V = V_A - V_B = 4\text{ V}$.

The electrostatic energy stored in the capacitor is given by:

$U = \dfrac{1}{2}CV^2$

$U = \dfrac{1}{2} \times 4\text{ }\mu\text{F} \times (4\text{ V})^2$

$U = 2 \times 16\text{ }\mu\text{J} = 32\text{ }\mu\text{J}$