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\(^{-2}\)]
A
1.8 x 10-10N
B
9.0 x 10-8N
C
9.0 x 102N
D
4.5 x 103N
Verified Correct Answer: C.
Physical Principles & Numerical Solution
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
Source: Myschool Past Questions Archive (Record #27660)