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. \(\frac{5}{9}q\)
• B. \(-\frac{5}{9}q\)
• C. \(-3q\)
• D. \(3q\)

Hint:

Coulomb's Law: \(F = k\frac{q_1q_2}{r^2}\)
Forces are vectors - consider direction
Net force zero means equal and opposite forces

Answer 1

The Correct Option is B

For net force on -2q to be zero: \[ F_{\text{by Q}} + F_{\text{by +5q}} = 0 \] \[ k\frac{Q(-2q)}{(r/3)^2} + k\frac{(+5q)(-2q)}{r^2} = 0 \] \[ \frac{9Q}{r^2} = \frac{10q}{r^2} \] \[ 9Q = 10q \Rightarrow Q = \frac{10q}{9} \] However, considering direction of forces: \[ \frac{kQ(2q)}{(r/3)^2} = \frac{k(5q)(2q)}{r^2} \] \[ \frac{18Q}{r^2} = \frac{10q}{r^2} \Rightarrow Q = \frac{5q}{9} \] But must be negative to oppose repulsion from +5q: \[ Q = -\frac{5q}{9} \]