Question

A given charge is situated at a certain distance from an electric dipole in the end-on position, experiences a force F. If the distance of the charge is doubled, the force acting on the charge will be

Moderate

Solution

We know that for end-on position
$\begin{aligned} E_{end on} = \frac{1}{4 {πε}_{0}} \times \frac{2 p}{r^{3}} \\ ∴ F = \frac{1}{4 {πε}_{0}} \times \frac{2 pq}{r^{3}} \end{aligned}$
When the distance is doubled, then
$\begin{array}{l} F^{'} = \frac{1}{4 {πε}_{0}} \times \frac{2 pq}{(2 r)^{3}} \\ = \frac{1}{8} [\frac{1}{4 {πε}_{0}} \times \frac{2 pq}{r^{3}}] = \frac{F}{8} \end{array}$

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