Two charges, each of \( 1\,\mu\mathrm{C} \), are placed on the x-axis at x = 1 m and x = 2 m. What is the magnitude of the net force on a charge of 1 C kept at the origin?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 11 250 N
By the superposition principle, the net force on the 1 C charge is the vector sum of the Coulomb forces of the two charges. Both charges are positive and lie on the same side (+x) of the origin. So both repel the 1 C charge along −x, and the two forces add.
The data in SI units are:
\( q=1\,\mu\mathrm{C}=1\times10^{-6}\,\mathrm{C} \)
\( Q=1\,\mathrm{C} \)
\( k=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \)
The force due to the charge at x = 1 m is:
\( \begin{aligned} F_1 &= \frac{kqQ}{r_1^{2}} \\ &= \frac{9\times10^{9}\times1\times10^{-6}\times1}{(1)^{2}}\,\mathrm{N} \\ &= 9000\,\mathrm{N} \end{aligned} \)
The force due to the charge at x = 2 m is:
\( \begin{aligned} F_2 &= \frac{kqQ}{r_2^{2}} \\ &= \frac{9\times10^{9}\times1\times10^{-6}\times1}{(2)^{2}}\,\mathrm{N} \\ &= 2250\,\mathrm{N} \end{aligned} \)
The two forces act in the same direction, so the net force is:
\( \begin{aligned} F &= F_1+F_2 \\ &= 9000\,\mathrm{N}+2250\,\mathrm{N} \\ &= 11\,250\,\mathrm{N} \end{aligned} \)