Three point charges lie on the x-axis: \(+2\ \mu\mathrm{C}\) at \(x=0\), \(+1\ \mu\mathrm{C}\) at \(x=30\ \mathrm{cm}\) and \(-4\ \mu\mathrm{C}\) at \(x=60\ \mathrm{cm}\). The magnitude of the net electrostatic force on the \(+1\ \mu\mathrm{C}\) charge is (Take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.6 N
The force on the \( +1\ \mu\mathrm{C} \) charge from each of the other charges is found by Coulomb’s law, as if the third charge were absent. The two forces are then added as vectors. Each of the other charges is 0.30 m away from it.
The \( +2\ \mu\mathrm{C} \) charge repels it, so this force points along \( +x \):
\( \begin{aligned} F_1 &= \frac{9\times10^{9}\times2\times10^{-6}\times1\times10^{-6}}{(0.30)^{2}} \\ &= 0.2\ \mathrm{N} \end{aligned} \)
The \( -4\ \mu\mathrm{C} \) charge attracts it, so this force also points along \( +x \):
\( \begin{aligned} F_2 &= \frac{9\times10^{9}\times4\times10^{-6}\times1\times10^{-6}}{(0.30)^{2}} \\ &= 0.4\ \mathrm{N} \end{aligned} \)
Both forces act in the same direction, so they add.
\( \begin{aligned} F &= F_1+F_2 \\ &= 0.2+0.4 \\ &= 0.6\ \mathrm{N} \end{aligned} \)
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 …”