Question

Three charges \(+5q\), \(Q\) and \(-2q\) are kept along a straight line in the same order such that, \(+5q\) and \(-2q\) charges are at a distance of \(\frac{2r}{3}\) and \(\frac{r}{3}\) from the charge \(Q\) respectively. If the net force on the charge \(-2q\) is zero, then \(Q\) is

• A. \(+\dfrac{5}{9}q\)
• B. \(-\dfrac{5}{9}q\)
• C. \(3q\)
• D. \(-3q\)

Hint:

For electrostatic equilibrium of a charge, first decide the directions of forces and then equate their magnitudes. The sign of the unknown charge is decided by whether the force must be attractive or repulsive.

Answer 1

The Correct Option is B

Step 1: Write forces on the \( -2q \) charge.
Distance from \( +5q \) to \( -2q \) is r. Force from \( +5q \): attractive, magnitude \( F_1 = k\frac{10q^2}{r^2} \) (toward \( +5q \)). Distance from Q to \( -2q \) is \( r/3 \). Force from Q: \( F_2 = k\frac{2q|Q|}{(r/3)^2} = \frac{18kq|Q|}{r^2} \).

Step 2: Set net force to zero.
For equilibrium, Q must repel \( -2q \), so Q is negative. Balancing magnitudes: \( 10kq^2/r^2 = 18kq|Q|/r^2 \Rightarrow |Q| = \frac{5q}{9} \), so \[ \boxed{Q = -\dfrac{5}{9}q} \]