Question

A point charge 'Q' is placed at the centre of the line joining two equal charges '+q' and '+q'. The value of 'Q' when the system is in equilibrium is

• A. \(\frac{+q}{4}\)
• B. \(\frac{+q}{2}\)
• C. \(\frac{-q}{2}\)
• D. \(\frac{-q}{4}\)

Hint:

For a +q charge to be in equilibrium, the force from the middle charge must cancel the force from the other +q.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Let the two +q charges be a distance \(2a\) apart, with Q at the centre, a distance \(a\) from each. For the whole system to be in equilibrium, every charge must have zero net force on it.

Step 2: Equilibrium of one end charge
Consider the right +q. The other +q repels it with force \(\dfrac{kq^2}{(2a)^2} = \dfrac{kq^2}{4a^2}\) to the right. The charge Q must pull it back with \(\dfrac{kqQ}{a^2}\), so Q must be negative:
\[ \frac{kq|Q|}{a^2} = \frac{kq^2}{4a^2} \Rightarrow |Q| = \frac q4 \]

Step 3: Check the middle charge
By symmetry, the forces on Q from the two +q charges cancel, so Q is also in equilibrium.
Therefore \(Q = -\dfrac q4\). Option (B) and (C) have magnitude q/2, which would over-pull the end charges, and (A) is positive, which would repel them further.

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