Q. The charge on two identical metallic balls are +40$\mu$ and $- 10\mu$C respectively and they are separated at 2.0 m. How much and nature of force will act between them?
A) 2.9 N, repulsive
B) 1.9 N, attractive
C) 1.2 N, repulsive
D) 0.9 N, attractive
Explanation
(a) According to Coulomb’s law $F = \frac{1}{4\pi\varepsilon_{0}}\frac{\mathcal{Q}_{1}\mathcal{Q}_{2}}{r^{2}\ }$
$= - \frac{9 \times 10^{9} \times 40 \times 10^{- 6} \times 10 \times 10^{- 6}}{(2.0)^{2}}$
$= - 0.9$ N
The force will be (attractive).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.