Question on: JAMB Physics - 1998
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
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Correct Option: 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
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