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
Ask EduPadi AI for a detailed answer
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
F =\(\frac {5 {\times} 10 {\times} 8 {\times} 10}{0.002 {\times} 0.002}\) x 9 x 10
F = 900N
Add your answer
Please share this, thanks!
No responses