A point charge Q is placed near two fixed point charges A and B. The net electrostatic force on Q is 0.5 N along the +x-axis. When charge A is removed, the force on Q becomes 0.2 N along the −x-axis. The force exerted on Q by charge A alone is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.7 N along the +x-axis
By the principle of superposition, the force of B on Q is the same whether A is present or not. The net force on Q is the vector sum of the force due to A and the force due to B.
Take the +x direction as positive. With A removed, only B acts, so \( F_B=-0.2\ \mathrm{N} \).
\( \begin{aligned} F_A &= F_{\text{net}}-F_B \\ &= (+0.5)-(-0.2) \\ &= +0.7\ \mathrm{N} \end{aligned} \)
Therefore, the force due to A alone is 0.7 N along the +x-axis.
NCERT (2026-27):
Forces Between Multiple Charges – “Experimentally, it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition.”