A point charge of \(+1\ \mu\mathrm{C}\) is at the origin. Charges of \(+3\ \mu\mathrm{C}\) and \(+4\ \mu\mathrm{C}\) are placed on the x-axis and the y-axis respectively, each at a distance of 10 cm from the origin. The magnitude of the net electrostatic force on the \(1\ \mu\mathrm{C}\) charge is (take \(k=9\times10^{9}\ \mathrm{N\,m^2\,C^{-2}}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4.5 N
By the principle of superposition, each force is found from Coulomb’s law and the two are added as vectors. Both forces are repulsive. One acts along the x-axis and the other along the y-axis, so they are at right angles.
Given:
\( Q=1\times10^{-6}\ \mathrm{C} \)
\( q_1=3\times10^{-6}\ \mathrm{C} \)
\( q_2=4\times10^{-6}\ \mathrm{C} \)
\( r=10\ \mathrm{cm}=0.1\ \mathrm{m} \)
\( \begin{aligned} F_1 &= \frac{kQq_1}{r^2} \\ &= \frac{9\times10^{9}\times1\times10^{-6}\times3\times10^{-6}}{(0.1)^2} \\ &= 2.7\ \mathrm{N} \end{aligned} \)
\( \begin{aligned} F_2 &= \frac{kQq_2}{r^2} \\ &= \frac{9\times10^{9}\times1\times10^{-6}\times4\times10^{-6}}{(0.1)^2} \\ &= 3.6\ \mathrm{N} \end{aligned} \)
\( \begin{aligned} F &= \sqrt{F_1^2+F_2^2} \\ &= \sqrt{(2.7)^2+(3.6)^2} \\ &= 4.5\ \mathrm{N} \end{aligned} \)
NCERT (2026-27):
Forces Between Multiple Charges – “… 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, …”
“The vector sum is obtained as usual by the parallelogram law of addition of vectors. …”