Question

A charge g is placed at the centre of the line joining the equal charges Q. The system of three charges will be in equilibrium if q is equal to

Easy

Solution

Let the distance between the two charges Q and Q be r. Force on Q due to charge g at a distance (r /2).
$= \frac{1}{4 {πε}_{0}} \times \frac{qQ}{(r / 2)^{2}}$
Force on Q due to other charge Q at a distance r
$\begin{array}{l} = \frac{1}{4 {πε}_{0}} \times \frac{Q \times Q}{(r)^{2}} \\ ∴ \frac{1}{4 {πε}_{0}} \times \frac{qQ}{(r / 2)^{2}} + \frac{1}{4 {πε}_{0}} \frac{Q \times Q}{r^{2}} = 0 \\ 4 q = - Q or q = - Q / 4 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME