Answer
(a) $E= 8000 \mathrm{V} / \mathrm{m} $
(b) $F = 19.2 \times 10^{-6} \mathrm{N}$
(c) The work done and the potential energy have the same magnitude.
Work Step by Step
(a) The electric field $E$ is related to the separated distance $d$ and the potential difference $V$ between the two plates
\begin{aligned} E &=\frac{V}{d} \\ &=\frac{360 \mathrm{V}}{0.045 \mathrm{m}} \\ &= \boxed{8000 \mathrm{V} / \mathrm{m} }
\end{aligned}
(b) The force $F$ exerted on the charge by
\begin{aligned}
F &=q E \\ &=\left(2.40 \times 10^{-9} \mathrm{C}\right)\left(8 \times 10^{3} \mathrm{V} / \mathrm{m}\right) \\ &= \boxed{19.2 \times 10^{-6} \mathrm{N}}
\end{aligned}
(c) The force exerted causes the charge to move a distance equal to $d$ and is given by
\begin{aligned}
W&=F d\\ &=\left(19.2 \times 10^{-6} \mathrm{N}\right)(0.045 \mathrm{m}) \\ &=\boxed{8.64 \times 10^{-7} \mathrm{J}}
\end{aligned}
(d) The change in the potential energy $\Delta U $ is given by
\begin{equation}
\begin{aligned} \Delta U &=-q \Delta V \\ &=-360 \mathrm{V}\left(2.40 \times 10^{-9} \mathrm{C}\right) \\ &=-8.64 \times 10^{-7} \mathrm{J} \end{aligned}
\end{equation}
The work done and the potential energy have the same magnitude.
$W = |U| = 8.64 \times 10^{-7} \mathrm{J}$
