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Q.
A charge $q$ is placed at the centre of line joining two equal point charges each equal to $+Q$. The system of $3$ charges will be in equilibrium if $q$ is equal to
KCETKCET 2003Electrostatic Potential and Capacitance
Solution:
The situation is as shown in the figure.
Let two equal charges Q each placed at points A and B at a distance r apart. C is the centre of AB where charge q is placed.
For equilibrium, net force on charge Q = 0
$ \therefore \frac{1}{\pi \varepsilon _0} \frac{1}{4\pi \varepsilon _0} \frac{Qq}{r/2}^2 = 0 $
$ \therefore \frac{1}{\pi \varepsilon _0} \frac{QQ}{r^2} + \frac{1}{4\pi \varepsilon _0} \frac{Qq}{(r/2)^2} = 0 $
$ \frac{1}{4\pi \varepsilon _0} \frac{Q^2}{r^2} = - \frac{1}{4\pi \varepsilon _0} \frac{4Qq}{r^2}\, \,$ or $\, \, Q = -4q\, \,$ or $\, \, q = -\frac{Q}{4} $
