Question

Three point charges \(4q\), \(Q\), and \(q\) are placed in a straight line of length \(L\) at points 0, \(L/2\), and \(L\) respectively. The net force on charge \(q\) is zero. The value of \(Q\) is:

• A. \(4q\)
• B. \(-q\)
• C. \(-0.5q\)
• D. \(-2q\)
• E. \(3q\)

Hint:

When forces are in opposite directions, set the magnitudes equal to each other to find the value of the unknown charge.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For the net force on the third charge (\(q\)) to be zero, the force exerted by the first charge (\(4q\)) and the second charge (\(Q\)) must be equal in magnitude but opposite in direction.
Step 2: Key Formula or Approach
Using Coulomb's Law: \( F = \frac{k q_1 q_2}{r^2} \). The net force on \(q\) at position \(L\) is: \[ F_{net} = F_{4q} + F_{Q} = 0 \]
Step 3: Detailed Calculation
1. Force due to \(4q\) (at distance \(L\)): \[ F_{4q} = \frac{k (4q)(q)}{L^2} \] 2. Force due to \(Q\) (at distance \(L/2\)): \[ F_{Q} = \frac{k (Q)(q)}{(L/2)^2} = \frac{4 k Q q}{L^2} \] 3. Set the sum to zero: \[ \frac{4kq^2}{L^2} + \frac{4kQq}{L^2} = 0 \] 4. Divide by common terms (\(\frac{4kq}{L^2}\)): \[ q + Q = 0 \implies Q = -q \]
Step 4: Final Answer
The value of Q is \( -q \).