Answer
$\sqrt\frac{3kQq}{W}$
where, k= $\frac{1}{4\pi\epsilon_{0}}$
Work Step by Step
For the vertical force of rod on bearing to be zero,
F(net upward)=F(net downward)
LHS: +Q applies an upward force of $\frac{kQq}{h^{2}}$ on +q
RHS: +Q applies an upward force of $\frac{2kQq}{h^{2}}$ on +2q
W is the downward force.
Hence, we can write:
$\frac{kQq}{h^{2}}$ + $\frac{2kQq}{h^{2}}$ = W
$\frac{3kQq}{h^{2}}$ = W
So, $h^{2}$ = $\frac{3kQq}{W}$
Thus, h = $\sqrt\frac{3kQq}{W}$
(where k= $\frac{1}{4\pi\epsilon_{0}}$)
