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: Option C
Explanation
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