Question

A positive charge of 5.0×10−5 C is 0.040 m from a second positive charge of 2.0×10−6 C . Calculate the force between the charges.(1 point) Responses 5.6×10−2 N 5.6×10−2 N 1.4×10−2 N 1.4×10−2 N 2.3×101 N 2.3×101 N 5.6×102 N

Answer 1

If two electrons that are apart get pushed toward each other, how does the repulsion between them change?(1 point) Responses Initial repulsion is high and decreases as they approach. Initial repulsion is high and decreases as they approach. Initial repulsion is low and decreases as they approach. Initial repulsion is low and decreases as they approach. Initial repulsion is high and increases as they approach. Initial repulsion is high and increases as they approach. Initial repulsion is low and increases as they approach.

Answer 2

The force between two charges can be calculated using Coulomb's Law:

F = ((k * q1 * q2) / r^2)

Where F is the force between the charges, k is the electrostatic constant (9 * 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.

Plugging in the given values:

F = ((9 * 10^9 N m^2/C^2) * (5.0 * 10^-5 C) * (2.0 * 10^-6 C)) / (0.040 m)^2

F = (9 * 10^9 * 5.0 * 2.0) / (10^-5 * 10^-6 * 0.040^2)

F = (9 * 5.0 * 2.0) / (10^-11 * 0.040^2)

F = (90 * 5.0 * 2.0) / (10^-11 * 0.0016)

F = (900 * 10^11) / (1.6 * 10^-12)

F = 562.5 * 10^23

F = 5.6 * 10^25 N

The correct answer is 5.6×10^25 N.

Answer 3

The correct answer is: Initial repulsion is high and decreases as they approach.

The repulsion between two electrons is due to their like charge. They both have a negative charge, so they repel each other. When they are far apart, the repulsion between them is high because the electrostatic force decreases with distance according to Coulomb's Law. As they are pushed closer together, the distance between them decreases, causing the repulsion to decrease.