Capacitors and resistors

In steady state we know that there is no current flowing through the capacitor.

We mark the point of $+8V$ as $A$ and the point of $+3V$ as $B$ and the earthed point as $C$

Let the current flowing from $A$ to $B$ be $i$

Net drop through $A$ to $B$ should be $5V$

$\Rightarrow 4i+i=5 V$ (I have added drop across the two resistors equated it to 5)

$\Rightarrow i=1 A$

Also we can write

${V}_{A}-{V}_{P}=4 V$

$\Rightarrow {V}_{P}=4 V$

In the circuit $PC$ no current is flowing hence the only drop occuring is due to the capacitor

Drop across $PC= {V}_{P}-{V}_{C}=4 V$

Drop across capacitor $= 4 V$

Energy in the capacitor $= \frac{1}{2}C{V}^{2}=24 \mu J$

1/2 CV2 AND V= 4 VOLT

Since a capacitor is connected between the point P and ground,after the full charging of the capacitor, no current will flow through that branch hence we can consider the 2 ohm resistor to be disconnected from the circuit. Hence the current(conventional) will flow from +8 V to +3 V,which will face the total resistance of(4+1=)5 ohms. so the potential drop at point P is (8-3) 4/5= 4 V therefore voltage across the capacitor will be 4 V and so the energy stored in the capacitor will be (3) 4*4/2=24 Microjoules

Imaginarily connect a cell of 8V at the 8V end and a cell of 2V at the 2V end,, since the grounded part has 0 voltage and in the circuit,, we can consider it equivalent to the other side of cell (negative terminal) so connect the ends of 8V and 2V imaginary cells to the other end of capacitor,,,

Now what you have is a parallel combination of cells whose EMF is given by (E1r2+E2r1)/(r1+r2) which turns out to be 4V, now energy in steady state is CV^2/2= 24 ujoules

(the relation can be derived, and it is derived from the more general khirchoff law)