The force of repulsion between two point positive charges 5\(\mu\)C and 8\(\mu\)C separated at a distance of 0.02m apart is. [\(\frac{1}{4{\pi}{\varepsilon}_o}\) = 9 x 10\(^9\)Nm\(^2\)C\(^{-2}\)]
- 1.8 x 10-10N
- 9.0 x 10-8N
-
9.0 x 102N
- 4.5 x 103N
The correct answer is: C
Explanation
F =\(\frac{1}{4{\pi}{\varepsilon}_o}\) x \(\frac {q_1q_2}{r^2}\)
F =\(\frac {5 {\times} 10^{-6} {\times} 8 {\times} 10^{-6}}{0.002 {\times} 0.002}\) x 9 x 10\(^9\)
F = 900N