(a) [tex]2.5\cdot 10^{-5} J[/tex]
The energy stored in a capacitor is given by:
[tex]U=\frac{1}{2}CV^2[/tex]
where
C is the capacitance
V is the potential difference across the capacitor
In this problem,
[tex]C=2.00 \mu F= 2.00\cdot 10^{-6}F[/tex]
V = 5.00 V
Substituting,
[tex]U=\frac{1}{2}(2.00\cdot 10^{-6})(5.00)^2=2.5\cdot 10^{-5} J[/tex]
(b) [tex]U'=5\cdot 10^{-5} J[/tex]
The capacitance of a parallel-plate capacitor is given by
[tex]C=\frac{\epsilon_0 A}{d}[/tex]
where [tex]\epsilon_0[/tex] is vacuum permittivity, A the area of the plates, d their separation. As we can see, the capacitance is inversely proportional to the distance d: since here the distance between the plates is doubled, therefore, the capacitance will be halved:
[tex]C' = \frac{C}{2}=\frac{2.00 \cdot 10^{-6}}{2}=1.00\cdot 10^{-6}F[/tex]
Now, we can rewrite the energy stored in the capacitor as
[tex]U' = \frac{1}{2}\frac{Q^2}{C'}[/tex]
where Q is the charge stored in the capacitor: since it has been disconnected by the battery, the charge stored has not changed, therefore the new energy stored will be simply twice the previous energy:
[tex]U' = \frac{1}{2}\frac{Q^2}{\frac{C}{2}} = 2 U[/tex]
And therefore,
[tex]U'=5\cdot 10^{-5} J[/tex]
c) [tex]U''=1.25\cdot 10^{-5} J[/tex]
Now the battery is reconnected, so the p.d. across the capacitor is again
V = 5.00 V
The energy stored is
[tex]U'' = \frac{1}{2}C'V^2[/tex]
where we said
[tex]C' = \frac{C}{2}[/tex]
Substituting,
[tex]U'' = \frac{1}{2}\frac{C}{2}V^2 = \frac{1}{2}U[/tex]
So, the energy is now half the initial energy, therefore
[tex]U''=1.25\cdot 10^{-5} J[/tex]
