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
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
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Coulomb's Law: \(F = k\frac{q_1q_2}{r^2}\)
Forces are vectors - consider direction
Net force zero means equal and opposite forces
Updated On: Mar 18, 2026
- \(\frac{5}{9}q\)
- \(-\frac{5}{9}q\)
- \(-3q\)
- \(3q\)
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The Correct Option is B
Solution and Explanation
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:
\[
Q = -\frac{5q}{9}
\]
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