Charge Balance Problem – Detailed Solution

Charge Balance in a Line

Problem: Three point charges are arranged along a straight line as follows:

• A charge \(4q\) is fixed at \(x=0\).
• A charge \(Q\) is fixed at \(x=\frac{l}{2}\).
• A charge \(q\) is fixed at \(x=l\).

Determine the value of \(Q\) so that the net electrostatic force on the charge \(q\) (at \(x=l\)) becomes zero.

Possible Answers:

• (A) \(Q=-q\)
• (B) \(Q=-2q\)
• (C) \(Q=-\frac{q}{2}\)
• (D) \(Q=4q\)

Select the Correct Option

(A) \(Q=-q\)

(B) \(Q=-2q\)

(C) \(Q=-\frac{q}{2}\)

(D) \(Q=4q\)

Step‑by‑Step Explanation

1. Step 1: Identify the forces on the charge \(q\) at \(x=l\). There are two contributions:
   - Force due to the charge \(4q\) at \(x=0\): The distance is \(l-0=l\). Using Coulomb’s law, \[ F_1 = k\frac{4q \cdot q}{l^2} = \frac{4kq^2}{l^2}. \] Since both \(4q\) and \(q\) (assumed positive) repel, this force is directed to the right.
2. Step 2: The force on \(q\) due to the charge \(Q\) at \(x=\frac{l}{2}\):
   The distance from \(Q\) to \(q\) is \(\ell - \frac{l}{2} = \frac{l}{2}\). The force is: \[ F_2 = k\frac{Q\cdot q}{\left(\frac{l}{2}\right)^2} = k\frac{Qq}{\frac{l^2}{4}} = \frac{4kQq}{l^2}. \] The direction of this force depends on the sign of \(Q\). To counterbalance the force \(F_1\) (which pushes \(q\) to the right), \(F_2\) must be directed to the left. Therefore, \(Q\) should be negative.
3. Step 3: For equilibrium (net force on \(q\) is zero), set \[ F_1 + F_2 = 0. \] Using directions (taking right as positive and left as negative): \[ \frac{4kq^2}{l^2} + \frac{4kQq}{l^2} = 0. \]
4. Step 4: Factor out common terms: \[ \frac{4kq}{l^2} \left(q + Q \right) = 0. \] Since \(\frac{4kq}{l^2}\) is nonzero, we must have: \[ q + Q = 0 \quad \Longrightarrow \quad Q = -q. \]
5. Final Answer: The net force on \(q\) will be zero if \(Q = -q\).