Find the location, magnitude and sign of the third charge.

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Solution

the third charge has to be of negative charge. Let it be $- Q$ . Let it be placed a t distance x from charge $q$
Therefore forces on charges $q$ ,
$\frac{1}{4 π ϵ} \frac{q 4 q}{L^{2}} = \frac{1}{4 π ϵ} \frac{q Q}{x^{2}}$

Force on charge $Q$ ,
$\frac{1}{4 π ϵ} \frac{q Q}{x^{2}} = \frac{1}{4 π ϵ} \frac{Q 4 q}{(L - x)^{2}}$
Force on charge $4 q$ ,
$\frac{1}{4 π ϵ} \frac{q 4 q}{L^{2}} = \frac{1}{4 π ϵ} \frac{Q 4 q}{(L - x)^{2}}$
Therefore we have ,
$\frac{1}{4 π ϵ} \frac{Q 4 q}{(L - x)^{2}} = \frac{1}{4 π ϵ} \frac{q Q}{x^{2}}$
$\Rightarrow \frac{4}{(L - x)^{2}} = \frac{1}{x^{2}}$
$\Rightarrow 4 x^{2} = L^{2} + x^{2} - 2 L x$
$\Rightarrow x = L / 3$
Using this value of x in
\frac{1}{4 \pi \epsilon} \frac{q4q}{L^2} =\frac{1}{4 \pi \epsilon} \frac{qQ}{x^2}
we get -Q = -4q/9

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