
$W_{e}+W_{g}=\frac{1}{2} m v^{2}$
We have work done by electrostatic force as
$W_{e}=q E l \sin \theta$
and work done by the gravitational force as
$W_{g}=m g(l-l \cos \theta)$
Thus we get
$q E l \sin \theta+m g(l-l \cos \theta)=\frac{1}{2} m v^{2}$
Thus we get
$m g \sin \theta+m g l-m g l \cos \theta=\frac{1}{2} m v^{2}$
as $\theta=45^{\circ},$ we get
$m g l=\frac{1}{2} m v^{2}$
also as $v=\omega l$
we get
$\omega=\sqrt{\frac{2 g}{l}}$

Let $C_1$ and $C_2$ be the capacitance of the system for $x =\frac{1}{3} d$ and $x =\frac{2 d }{3}$, respectively. If $C _1=2 \mu F$ the value of $C _2$ is $........... \mu F$


