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(B) Let the two charges $Q$ be placed at points $A$ and $B$ separated by a distance $2r$. The charge $q$ is placed at the midpoint $O$.For the system

Vedclass · Accepted Answer

$-\frac{Q}{4}$

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(B) Let the two charges $Q$ be placed at points $A$ and $B$ separated by a distance $2r$. The charge $q$ is placed at the midpoint $O$.For the system to be in equilibrium,the net force on each charge must be zero.Consider the force on charge $Q$ at point $A$:The force due to charge $q$ at $O$ is $F_1 = \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{r^2}$ (attractive,towards $O$).The force due to charge $Q$ at $B$ is $F_2 = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{(2r)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{4r^2}$ (repulsive,away from $B$).For equilibrium,$F_1 + F_2 = 0$,so $\frac{1}{4 \pi \varepsilon_0} \frac{Qq}{r^2} + \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{4r^2} = 0$.Solving for $q$: $\frac{Qq}{r^2} = -\frac{Q^2}{4r^2} \Rightarrow q = -\frac{Q}{4}$.

$A$ charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to:

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