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: Understanding the Concept:
Take two charges +q at a separation L, with Q at the midpoint. The system is in equilibrium if each charge feels zero net force. By symmetry it is enough to check one +q.

Step 2: Forces on the right-hand +q:
Repulsion from the other +q at distance L: \(F_1 = \frac{kq^2}{L^2}\) (outward).
Force from Q at distance \(\frac L2\): \(F_2 = \frac{kqQ}{(L/2)^2} = \frac{4kqQ}{L^2}\). To balance, this force must point inward, so Q must be opposite in sign to q (attractive).

Step 3: Balance:
\[ \frac{kq^2}{L^2} = \frac{4k\,q\,|Q|}{L^2} \Rightarrow |Q| = \frac q4 \]
So \(Q = -\frac q4\). A positive Q would push the end charges outward, so options (A) and (B) cannot give equilibrium, and \(-\frac q2\) gives too large an attraction.

Final Answer:
The charge at the centre is \(-\frac q4\), option (D). \[ \boxed{-\frac{q}{4}} \]