Question

A charge 'q' placed at the centre on the line joining two charges 'Q' will be in equilibrium if 'q' will be equal to

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

Hint:

For a system of charges to be in equilibrium, the net force on \textit{every} charge must be zero. Don't just check the equilibrium for the central charge, which is a common mistake. The stability of the outer charges is what determines the value of the central charge.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The question asks for the value of charge 'q' such that the entire system of three charges is in equilibrium. This means the net force on each of the three charges must be zero. A charge 'q' placed at the center between two equal charges 'Q' will always be in equilibrium by symmetry. The crucial condition is to ensure that the charges 'Q' are also in equilibrium.
Step 2: Key Formula or Approach:
Let the two charges 'Q' be placed at \(x = -a\) and \(x = +a\). The charge 'q' is at the center, \(x = 0\). For the system to be in equilibrium, the net force on the charge Q at \(x = +a\) must be zero. \[ \vec{F}_{\text{net on Q at +a}} = \vec{F}_{\text{due to Q at -a}} + \vec{F}_{\text{due to q at 0}} = 0 \] Step 3: Detailed Explanation:
The force exerted by the charge Q at \(x=-a\) on the charge Q at \(x=+a\) is: \[ F_{QQ} = k \frac{Q \cdot Q}{(2a)^2} = k \frac{Q^2}{4a^2} \] This force is repulsive and acts along the +x direction.
The force exerted by the charge q at \(x=0\) on the charge Q at \(x=+a\) is: \[ F_{qQ} = k \frac{q \cdot Q}{a^2} \] For the net force on Q to be zero, this force must be equal in magnitude and opposite in direction to \(F_{QQ}\). Therefore, \(F_{qQ}\) must be attractive, which implies that q and Q must have opposite signs. So, q must be negative.
Setting the sum of forces to zero: \[ k \frac{Q^2}{4a^2} + k \frac{qQ}{a^2} = 0 \] \[ k \frac{qQ}{a^2} = -k \frac{Q^2}{4a^2} \] Cancel \(k, Q,\) and \(a^2\) from both sides: \[ q = -\frac{Q}{4} \] Step 4: Final Answer:
For the entire system of three charges to be in equilibrium, the central charge q must be equal to \(-Q/4\). Option (B) is correct.