Answer
(a) $C = 9.27 \times 10^{-9} \mathrm{~F}$
(b) $V = 2 \times 10^{9} \,\text{V}$
(c) $E = 7.2 \times 10^{5} \,\text{V/m} $
(d) $U = 2\times 10^{10} \,\text{J}$
Work Step by Step
(a) The capacitance depends on the area $A$ of the plate and the distance between the two plates and it is given by
$$C = \dfrac{ \epsilon_o A}{d}$$
Substitute by the values of $\epsilon_o, A$ and $d$ to get the capacitance $C$
\begin{align*}
C &= \dfrac{ \epsilon_o \pi r^2 }{d} \\
&= \dfrac{ (8.85 \times 10^{-12} \,\text{F/m}) \pi (3000 \,\text{m})^2 } {3 \times 10^{3} \,\text{m}} \\
& = \boxed{9.27 \times 10^{-9} \mathrm{~F}}
\end{align*}
(b) The capacitance between the two plates occur due to the accumulation of the charges on the plates and the potential is given by
$$V_{ab} = \dfrac{Q}{C} = \dfrac{20 \,\text{C}}{9.27 \times 10^{-9} \,\text{F}} =\boxed{ 2 \times 10^{9} \,\text{V}}$$
(c) The electric field is given by
$$E = \dfrac{Q}{A \epsilon_o} = \dfrac{20 \,\text{C}}{ \pi (3000 \,\text{m})^2 (8.85 \times 10^{-12} \,\text{F/m})} = \boxed{ 7.2
\times 10^{5} \,\text{V/m}}$$
(d) Energy $U$ is the amount of work stored in the capacitor and it is related to the charges on the plates by
$$U = \dfrac{1}{2} QV = \dfrac{1}{2} (20 \,\text{C})( 2 \times 10^{9}\,\text{V}) = \boxed{2\times 10^{10} \,\text{J}}$$
