Answer :
Explanation:
(a) It is known that energy present in electric field is the sum of energies of the capacitors. Since, there are only two capacitors so net energy of the capacitor will be as follows.
U = [tex]U_{1} + U_{2}[/tex]
= [tex]\frac{q^{2}_{1}}{2C_{1}} + \frac{q^{2}_{2}}{2C_{2}}[/tex]
As the charge is equally distributed, so [tex]q_{1} = q_{2} = \frac{q_{o}}{2}[/tex]
Also, potential difference of both of them is equal.
So, [tex]V_{1} = V_{2}[/tex]
or, [tex]\frac{q_{1}}{C_{1}} = \frac{q_{2}}{C_{2}}[/tex]
[tex]\frac{q_{o}}{2C_{1}} = \frac{q_{o}}{2C_{2}}[/tex]
as [tex]C_{1} = C_{2} = C[/tex]
U = [tex]\frac{q^{2}_{o}}{8C} + \frac{q^{2}_{o}}{8C}[/tex]
= 2 J
Thus, we can conclude that if the charge distributes equally, then total energy stored in the electric fields is 2 J.