Two point charges are separated by a distance r and repel each other with forces of magnitude F.

Two point charges are separated by a distance r and repel each other with forces of magnitude F. If their separation is reduced to 0.25 times the original value, what is the magnitude of the forces of repulsion?

Expert Solution

Calculate the magnitude of the force of repulsion between the two point charges using the formula.

Use the coulombs law, and then the magnitude of the force of attraction repulsion between the point charges is as follows:

F = kq1q2/r2

Here, k is electrostatic constant, q1 and q2 are two point charges, and r is the distance between the two point charges.

When the separation between the two point charges are reduced to a new distance r, then the new force of repulsion between the two point charges is as follows:

F = kq1q2/(r)2

The reduced distance between two point charges is equal to 0.25 times of the original distance between them.

r = 0.25 r

Substitute 0.25r for r in the equation F = kq1q2/(r)2 and solve for F.

F = kq1q2/(0.25r)2

= {1/(0.25)2}(kq1q2/r2)

= 16(kq1q2/r2)

Substitute F for kq1q2/r2.

F = 16 F

Therefore, the magnitude of the force of repulsion between the two charges increases 16 times to its original value.

Umair answered 1 year ago

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