c
In equilibrium, $\mathrm{F}_{\mathrm{e}}=\mathrm{T} \sin \theta$
$\mathrm{mg}=\mathrm{T} \cos \theta$
$\tan \theta=\frac{F_{e}}{m g}=\frac{q^{2}}{4 \pi \in_{0} x^{2} \times m g}$
$\therefore x = \sqrt {\frac{{{q^2}}}{{4\pi {\varepsilon _0}\,\tan \,\theta \,mg}}} $
Electric potential at the centre of the line
$V=\frac{k q}{x / 2}+\frac{k q}{x / 2}=4 \sqrt{k m g / \tan \theta}$
