If the distance between 2 charged particles is double, the force between them changes by a factor of?

Physics

1 answer:

aev [14]3 years ago

3 0

Answer:

See explanation

Explanation:

Given that;

q1 and q2 are the magnitudes of the two charges while r1 is the distance between the charges.

F1 =kq1q2/r1^2

Then,

F2 = kq1q2/(2r1)^2

F2 = kq1q2/4r1^2

So,

F2 = 1/4 F1

The force between the charges changes by a factor of 1/4.

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