Three charges $($ two of same nature $)$ are arranged on a straight line are in equilibrium. This nature of equilibrium is

stable

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Any of the above

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unstable

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neutral

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Solution

The correct option is C unstable
FBD of the charges, if charge $(- Q_{3})$ is slightly displaced to right.

Here,
$F_{1}$ is electrostatic force between charges $Q_{2} & Q_{3}$
$F_{2}$ is electrostatic force between charges $Q_{1} & Q_{3}$
$F_{3}$ is electrostatic force between charges $Q_{2} & Q_{1}$
$F_{1} = \frac{k Q_{2} Q_{3}}{(r_{2} - Δ r)^{2}} > F$ ........ $(1)$
$F_{2} = \frac{k Q_{1} Q_{3}}{(r_{2} + Δ r)^{2}} < F$ ........ $(2)$
$F_{3} = \frac{k Q_{2} Q_{1}}{(r_{1} + r_{2})^{2}} = F$ ........ $(3)$

From $(1), (2)$ and $(3)$ ,
$F_{1} \neq F_{2} \neq F_{3}$

For charge $+ Q_{1}$ ,
${\to F}_{n e t} = {\to F}_{3} - {\to F}_{2} \neq 0$

For charge $+ Q_{2}$ ,
${\to F}_{n e t} = {\to F}_{1} - {\to F}_{3} \neq 0$

For charge $+ Q_{3}$ ,
${\to F}_{n e t} = {\to F}_{1} - {\to F}_{2} \neq 0$
It cannot return to original position.

Hence, the charges are in unstable equilibrium.

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