A charge Q is placed at the centre of the line joining two equal charges q. Show that the system of three charges will be in equilibrium if $Q = \frac{- q}{4}$ .

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Solution

Let us consider the problem:
Let's see the force on one $Q$ .
$F = \frac{k Q^{2}}{4 a^{2}} + \frac{k q Q}{a^{2}}$
For the system to be equilibrium $F = 0$ .
$\frac{k Q^{2}}{4 a^{2}} + \frac{k q Q}{a^{2}} = 0$
$q = - \frac{Q}{4} .$

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A charge Q is placed at the centre of the line joining two equal charges q. Show that the system of three charges will be in equilibrium if Q=−q4.

Let us consider the problem:Let's see the force on one Q.F=kQ24a2+kqQa2For the system to be equilibrium F=0.kQ24a2+kqQa2=0q=−Q4.

A charge Q is placed at the centre of the line joining two equal charges q. Show that the system of three charges will be in equilibrium if $Q = \frac{- q}{4}$ .

Let us consider the problem:
Let's see the force on one $Q$ .
$F = \frac{k Q^{2}}{4 a^{2}} + \frac{k q Q}{a^{2}}$
For the system to be equilibrium $F = 0$ .
$\frac{k Q^{2}}{4 a^{2}} + \frac{k q Q}{a^{2}} = 0$
$q = - \frac{Q}{4} .$