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 ?

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 ?

(a)  \( \dfrac{Q}{2} \)

(b) \( -\dfrac{Q}{4} \)

(c) \( \dfrac{Q}{4} \)

(d) \( -\dfrac{Q}{2} \)

Answer :  (b) \( -\dfrac{Q}{4} \)

Concept: If q is +ve then the forces acting on Q will both be repulsion and it cannot be in equilibrium. So the forces acting on Q has to be equal and opposite, which is possible only when q is -ve.

\( F_1 = \dfrac{k Q Q}{(2a)^2} \)

\( F_2 = \dfrac{k Q q}{a^2} \)

\( F_1 = F_2 \quad \text{For equilibrium} \)

\( \dfrac{k Q Q}{4a^2} = \dfrac{k Q q}{a^2} \)

\( q = \dfrac{Q}{4} \)

\( \text{For Stable equilibrium q has to be -ve.} \)