$U _{ f }=\frac{ kq P }{\left(2 \ell \sin \frac{\alpha}{2}\right)^2}+ mgh$

Now, from $\triangle OAB$

$\alpha+90-\theta+90-\theta=180$

$\Rightarrow \alpha=2 \theta$

From $\triangle ABC : h =2 \ell \sin \left(\frac{\alpha}{2}\right) \sin \theta$

$h =2 \ell \sin \left(\frac{\alpha}{2}\right) \sin \left(\frac{\alpha}{2}\right)$

$\Rightarrow  h =2 \ell \sin ^2\left(\frac{\alpha}{2}\right)$

Now charge is in equilibrium at point $B$.

So, using sine rule

$\Rightarrow \frac{ mg }{\sin \left[90+\frac{\alpha}{2}\right]}=\frac{ qE }{\sin [180-2 \theta]}$

$\Rightarrow \frac{ mg }{\cos \frac{\alpha}{2}}=\frac{ qE }{\sin 2 \theta}$

$\Rightarrow \frac{ mg }{\cos \frac{\alpha}{2}}=\frac{ qE }{\sin \alpha} \Rightarrow \frac{ mg }{\cos \frac{\alpha}{2}}=\frac{ qE }{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}$

$\Rightarrow qE = mg 2 \sin \left(\frac{\alpha}{2}\right)$

$\Rightarrow \frac{ q 2 lq }{\left[2 \ell \sin \frac{\alpha}{2}\right]^3}= mg 2 \sin $$\left(\frac{\alpha}{2}\right) \Rightarrow \frac{ kpq }{\left[2 \ell \sin \frac{\alpha}{2}\right]^2}= mg \sin \left(\frac{\alpha}{2}\right) \times\left(2 \ell \sin \frac{\alpha}{2}\right)$

$\Rightarrow \frac{ kpq }{\left[2 \ell \sin \frac{\alpha}{2}\right]^2}= mgh \Rightarrow \operatorname{substituting~this~in~equation~(i)~}$

$\Rightarrow U$

$W =\Delta U = Nmgh = N =2$