Two equal charges of the same sign are placed on a straight line. A third charge, equal in magnitude but opposite in sign to the other two, is placed midway between them. The system will
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → not be in equilibrium
A system of charges is in equilibrium only when the net force on every charge is zero. The two end charges pull the middle charge equally, so it is balanced. The answer depends on the force on each end charge.
Let the end charges be \( Q \), a distance \( 2a \) apart, with \( -Q \) midway. By Coulomb’s law, the repulsion of the other end charge on one end charge is:
\( \begin{aligned} F_{\mathrm{rep}} &= \frac{kQ\cdot Q}{(2a)^{2}} \\ &= \frac{kQ^{2}}{4a^{2}} \end{aligned} \)
The attraction of the middle charge on it is:
\( \displaystyle F_{\mathrm{att}}=\frac{kQ^{2}}{a^{2}} \)
From the above, \( F_{\mathrm{att}}=4F_{\mathrm{rep}} \). Hence each end charge feels a net force \( 3kQ^{2}/4a^{2} \) towards the middle charge.
Option (1) be in equilibrium — Incorrect. Only the middle charge is balanced. Each end charge is pulled towards the centre.
Option (2) not be in equilibrium — Correct. The nearer middle charge attracts it four times as strongly as the farther end charge repels it. So the net force on each end charge is not zero.
Option (3) lose charge — Incorrect. The forces between the charges can move them but cannot remove charge from them. By conservation of charge, the total charge of the isolated system stays the same.
Option (4) oscillate — Incorrect. The net force on each end charge always points towards the middle charge, so the end charges move in towards it. No restoring force acts about a fixed position, so they cannot oscillate.