Question

Two point charges +e and +4e are kept at a distance 'd' units apart. Third point charge +q is placed between the two charges at a distance 'x' units from charge +e so as to be in equilibrium. The value of 'x' is

• A. \(\frac{d}{3}\) units
• B. \(\frac{d}{2}\) units
• C. \(\frac{2d}{5}\) units
• D. \(\frac{3d}{4}\) units

Hint:

Force on one end charge from the other end charge must balance the pull from the middle charge.

Answer 1

The Correct Option is A

Step 1: Use potential energy idea:
For equilibrium of the end charge, the net force must be zero. Let the separation of the end charges be $2d$, so Q is at distance $d$ from each.

Step 2: Equate magnitudes:
Repulsion between ends: $\frac{kq^2}{(2d)^2} = \frac{kq^2}{4d^2}$. Attraction to centre: $\frac{kq|Q|}{d^2}$. Equal when $|Q| = \frac q4$, with the opposite sign to q. So $Q = -\frac q4$ (D).

Final Answer:
$-\frac q4$. \[ \boxed{-\frac{q}{4}} \]