Answer
x = $\frac{L}{2}$(1 + $\frac{kQq}{Wh^2}$)
{where, k = $\frac{1}{4\pi\epsilon_0}$}
Work Step by Step
The rod is balanced. Hence, net torque ($\tau$) = 0
So, $\tau$ (clockwise) = $\tau$ (anti-clockwise)
Here, F1 and W are applying torque in a clockwise direction, whereas, F2 is applying torque in an anti-clockwise direction...... (1)
We know that, $\tau$ = rFsin$\theta$...... (2)
where r = distance from pivot point.
Torque caused by F1 = ($\frac{kQq}{h^2}$)($\frac{L}{2}$)
= $\frac{kQqL}{2h^2}$ clockwise ......(3)
Torque caused by F2 = ($\frac{2kQq}{h^2}$)($\frac{L}{2}$)
= $\frac{kQqL}{h^2}$ anti-clockwise ......(4)
Torque caused by W = W(x-$\frac{L}{2}$)
= $\frac{W}{2}$(2x-L) clockwise ......(5)
Using (1), (3), (4) and (5):
$\frac{kQqL}{2h^2}$ + $\frac{W}{2}$(2x-L) = $\frac{kQqL}{h^2}$
$\frac{W}{2}$(2x-L) = $\frac{kQqL}{2h^2}$
2x-L = $\frac{kQqL}{Wh^2}$
2x = L + $\frac{kQqL}{Wh^2}$
Thus, x = $\frac{L}{2}$(1 + $\frac{kQq}{Wh^2}$)
{where, k = $\frac{1}{4\pi\epsilon_0}$}
