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Principle of Superposition

Point charges...

Question

Point charges $q_{1}$ and $q_{2}$ lie on the $x - a x i s$ at points $x = - a$ and $x = + a$ respectively. Given the net electrostatic force on point charge $+ Q$ placed at $x = + a / 2$ to be zero and $q_{1} = n q_{2}$ . Find $n$ .

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Solution

For the force on $Q$ to be zero,

$\frac{1}{4 π ϵ_{0}} \frac{q_{1} Q}{(3 a / 2)^{2}} = \frac{1}{4 π ϵ_{0}} \frac{q_{2} Q}{(a / 2)^{2}}$

Therefore,
$\frac{q_{1}}{9} = q_{2}$

$\Rightarrow q_{1} = 9 q_{2}$

$\Rightarrow n = 9$

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