Three charges, \( +Q \), \( q \) and \( +Q \) are placed with \( q \) at mid point of the line joining the other two charges. For the system of charges to be in equilibrium, the value of \( q \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( -Q/4 \)
The middle charge \( q \) is pulled equally from both sides, so it is in equilibrium for any value. The condition comes from an end charge: the net Coulomb force on it must be zero.
Let the end charges be a distance \( 2d \) apart, with \( q \) at distance \( d \) from each. The force on the right-hand \( +Q \) due to the left-hand \( +Q \) (repulsion) and due to \( q \) must add to zero:
\( \displaystyle \frac{kQ\cdot Q}{(2d)^{2}} + \frac{kQ\,q}{d^{2}} = 0 \)
\( \displaystyle \frac{Q}{4} + q = 0 \;\Rightarrow\; q = -\frac{Q}{4} \)
The middle charge must be negative, so that its attraction cancels the repulsion between the two end charges.