An insulating long light rod of length L is pivoted at its center O and balanced with a weight W at a distance x from the left end as shown in figure. Charges q and 2q are fixed at the two ends of the rod. Exactly below each of these charges at a distance h a positive charge Q is fixed. If the rod is in equilibrium, then x is

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An insulating long light rod of length L is pivoted at its center O and balanced with a weight W at a distance x from the left end as shown in figure. Charges q and 2q are fixed at the two ends of the rod. Exactly below each of these charges at a distance h a positive charge Q is fixed. If the rod is in equilibrium, then x is

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$\frac{4 QLq + ε_{0} h^{2} LW}{8 {πh}^{2} W}$

$\frac{QLq + ε_{0} h^{2} LW}{ε_{0} h^{2} W}$

$\frac{QLq + 4 {πε}_{0} h^{2} LW}{8 {πε}_{0} h^{2} W}$

$\frac{QLq + ε_{0}^{2} h^{2} LW}{h^{2} W}$

answer is D.

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Detailed Solution

By Law of Conservation of Moments about O,
$(\begin{array}{l} Total Clokwise \\ (CW) Moments \end{array}) = (\begin{array}{l} Total Counter \\ Clockwise (CCW) \\ Moments \end{array})$
$\Rightarrow \frac{qQ}{4 {πε}_{0} h^{2}} (\frac{L}{2}) + W (x - \frac{L}{2}) = \frac{2 qQ}{4 {πε}_{0} h^{2}} (\frac{L}{2})$
$\Rightarrow W (x - \frac{L}{2}) = \frac{QqL}{8 {πε}_{0} h^{2}} \Rightarrow x = \frac{QqL + 4 {πε}_{0} h^{2} LW}{8 {πε}_{0} h^{2} W}$