The force of repulsion between two point positive charges 5\(\mu\)C and 8\(\mu\)C separated at a distance of 0.02 map art is. [\(\frac{1}{4{\pi}{\varepsilon}_o}\) = 9 x 109Nm2C-2]

A.

1.8 x 10-10N

B.

9.0 x 10-8N

C.

9.0 x 102N

D.

4.5 x 103N

Correct answer is C

F =\(\frac{1}{4{\pi}{\varepsilon}_o}\) x \(\frac {q_1q_2}{r^2}\)

F =\(\frac {5 {\times} 10 {\times} 8 {\times} 10}{0.002 {\times} 0.002}\) x 9 x 10

F = 900N