Question

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

• A. stable
• B. unstable
• C. neutral
• D. Any of the above

Answer 1

The correct option is B 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 $Q_2\&Q_1$
$F_1=\frac{k{Q}_2{Q}_3}{(r_2-\Delta r{)}^2}>F$ ........$\left(1\right)$
$F_2=\frac{k{Q}_1{Q}_3}{(r_2+\Delta r{)}^2}<F$ ........$\left(2\right)$
$F_3=\frac{k{Q}_2{Q}_1}{(r_1+r_2{)}^2}=F$ ........$\left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$,
$F_1\ne F_2\ne F_3$

For charge $+Q_1$,
${\overrightarrow{F}}_{net}={\overrightarrow{F}}_3-{\overrightarrow{F}}_2\ne 0$

For charge $+Q_2$,
${\overrightarrow{F}}_{net}={\overrightarrow{F}}_1-{\overrightarrow{F}}_3\ne 0$

For charge $+Q_3$,
${\overrightarrow{F}}_{net}={\overrightarrow{F}}_1-{\overrightarrow{F}}_2\ne 0$
It cannot return to original position.

Hence, the charges are in unstable equilibrium.