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antoniya [11.8K]
3 years ago
10

A -0.00125 C charge is placed 3.62 m from a +0.00333 C charge. What is the magnitude of the electric force between them?

Physics
1 answer:
olga55 [171]3 years ago
7 0

Answer: 2858.7726N

Explanation:

Given that :

Charge 1 (q1) = -0.00125 C

Charge 2 (q2) = +0.00333 C

Distance (r) = 3.62 m

According to columb's Law: The magnitude of the force between any two charged objects :

Fe = (Kq1q2) / r^2

Coloumb's constant(k) = 9×10^9 Nm^2/C^2

Fe = (9×10^9 * 0.00125 * 0.00333) / 3.62^2

Fe = (0.0000374625 * 10^9) / 13.1044

Fe = 37462.5 / 13.1044

Fe = 2858.7726N

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