Three point charges, each \( +q \), are placed on a straight line, with a separation \( a \) between neighbouring charges. The magnitude of the net electrostatic force on one of the end charges is (\( k=1/4\pi\varepsilon_0 \))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \frac{5kq^{2}}{4a^{2}} \)
By the principle of superposition, the force between any two charges follows Coulomb’s law, as if the other charges were absent. The net force is the vector sum of these forces.
Call the charges A, B and C in order, with A at one end. B is at a distance \( a \) from A and C at \( 2a \). Both are positive, so both repel A. Their forces on A act in the same direction, along the line and away from B and C. The two forces are:
\( \displaystyle F_{AB}=\frac{kq^{2}}{a^{2}} \)
\( \begin{aligned} F_{AC} &= \frac{kq^{2}}{(2a)^{2}} \\ &= \frac{kq^{2}}{4a^{2}} \end{aligned} \)
As they act along the same direction, they add:
\( \begin{aligned} F &= F_{AB}+F_{AC} \\ &= \frac{kq^{2}}{a^{2}}+\frac{kq^{2}}{4a^{2}} \\ &= \frac{5kq^{2}}{4a^{2}} \end{aligned} \)
NCERT (2026-27):
Forces Between Multiple Charges – “The principle of superposition says that in a system of charges q1, q2, ..., qn, the force on q1 due to q2 is the same as given by Coulomb’s law, i.e., it is unaffected by the presence of the other charges q3, q4, ..., qn. The total force F1 on the charge q1, due to all other charges, is then given by the vector sum of the forces F12, F13, ..., F1n …”