Answer
(a) $C = 7.85 \times 10^{4} \dfrac{\mathrm{V}}{ \mathrm{m}^{4 / 3}}$
(b) $-\frac{4}{3}\left(7.85 \times 10^{4} \dfrac{\mathrm{V}}{ \mathrm{m}^{4 / 3}}\right) x^{1 / 3}$
(c) $F = 3.13 \times 10^{-15} \mathrm{N}$
Work Step by Step
(a) The value of $C$ is calculated by
\begin{aligned}
C &=\frac{V}{x^{4/3}} \\ &=\frac{240 \mathrm{V}}{(0.013 \mathrm{m})^{4 / 3}} \\
&=\boxed{7.85 \times 10^{4} \dfrac{\mathrm{V}}{ \mathrm{m}^{4 / 3}}} \end{aligned}
(b) The electric field equals the differentiation of the potential to the distance
\begin{align}
E& =-\frac{\partial V}{\partial x}\\
&=-\frac{\partial}{\partial x}\left(C x^{4 / 3}\right)\\
&=-\frac{4}{3} C x^{1 / 3}\\
&=-\frac{4}{3}\left(7.85 \times 10^{4} \dfrac{\mathrm{V}}{ \mathrm{m}^{4 / 3}}\right) x^{1 / 3}
\end{align}
(c) At the half way, the distance will be $x$ = 13 mm /2 = 6.5 mm and the force will be calculated by
\begin{aligned}
F &= qE \\
&=\left(-1.60 \times 10^{-19} \mathrm{C}\right) \times -\frac{4}{3}\left(7.85 \times 10^{4} \dfrac{\mathrm{V}}{ \mathrm{m}^{4 / 3}}\right) (0.0065 \mathrm{m})^{1 / 3} \\
&=\boxed{3.13 \times 10^{-15} \mathrm{N}}
\end{aligned}
