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Two like and equal charges are at a distance of d=5cm and exert a force of F=9×10 ^−3

N on each other. What is the magnitude of each charge? What is the direction of the electrostatic force between them?
a) 4.5×10^ −3C, attractive
b)9×10−^3C, repulsive
c) 18×10^−3C, attractive
d) 5×10^ −3C, repulsive

1 Answer

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Final answer:

The magnitude of each charge is 4.5×10−3 C, and the direction of the electrostatic force between them is repulsive. Coulomb's law was used to calculate the charges based on the given force and distance information.

Step-by-step explanation:

To determine the magnitude of each charge that exerts a force of F = 9×10−3 N when separated by a distance of d = 5 cm, we use Coulomb's law, which is stated as:

F = k * |q1 * q2| / d2

where k is Coulomb's constant (8.988 × 109 N·m2/C2), and q1 and q2 are the magnitudes of the charges. Since the

charges are equal, we can say q = q1 = q2.

Plugging in the values and solving for q we get:

q = √(F * d2 / k)

q = √(9×10−3 N * (0.05 m)2 / (8.988 × 109 N·m2/C2))

q ≈ √(2.245 × 10−10 C2)

q ≈ 4.5×10−3 C

Since both charges are like and equal, the electrostatic force between them will be repulsive. Therefore, option (d) 5×10−3C, repulsive, is the correct answer.

User Usmanwalana
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