Point charges of \(+1\ \mu\mathrm{C}\) each are fixed at the points (−30 cm, 0) and (30 cm, 0). A point charge of \(+2\ \mu\mathrm{C}\) is fixed at the point (0, 30 cm). The magnitude of the net electrostatic force on a charge of \(+1\ \mu\mathrm{C}\) placed at the origin 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) → \(0.2\ \mathrm{N}\)
Each fixed charge exerts its own Coulomb force on the charge at the origin. The net force is the vector sum of these forces. All three charges are 30 cm from the origin.
Force due to each \( +1\ \mu\mathrm{C} \) charge on the x-axis:
\( \begin{aligned} F_1 &= \frac{1}{4\pi\varepsilon_0}\,\frac{q_1q_0}{r^{2}} \\ &= \frac{9\times10^{9}\times10^{-6}\times10^{-6}}{(0.3)^{2}} \\ &= 0.1\ \mathrm{N} \end{aligned} \)
These two forces are repulsive. One acts along \( +x \) and the other along \( -x \), so their sum is zero.
Force due to the \( +2\ \mu\mathrm{C} \) charge:
\( \begin{aligned} F_2 &= \frac{9\times10^{9}\times2\times10^{-6}\times10^{-6}}{(0.3)^{2}} \\ &= 0.2\ \mathrm{N} \end{aligned} \)
Therefore, the net force on the charge at the origin is \( 0.2\ \mathrm{N} \), directed along the negative y-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.”