Three point charges 4q, Q and q are placed in a straight line of length \( l \) as shown. The net force on charge q is zero when value of q is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → −Q
By Coulomb's law the force on the charge at the right end is the vector sum of the forces from 4q and from Q. Both lie on the same line, so the net force is zero only when the two forces are equal and opposite. From the figure, 4q is at distance \( l \) from q, and Q is at distance \( l/2 \) from q.
\( \displaystyle F_{\text{net}} = \frac{1}{4\pi\varepsilon_0}\left[\frac{q\cdot4q}{l^{2}} + \frac{q\cdot Q}{(l/2)^{2}}\right] = 0 \)
\( \displaystyle \frac{4q^{2}}{l^{2}} + \frac{4qQ}{l^{2}} = 0 \)
\( 4q(q + Q) = 0 \Rightarrow q = -Q \)