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Two charges q1 and q2 are separated by a distance r and apply a force F to each other. If both charges are doubled, and the distance between them is halved, the new force between them is

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Lanuel

Answer:

The new force between them is increased by a factor of 16.

Explanation:

According to Coulombs law, the force of attraction between two (2) charges is given by the formula;

F = Kq1q2/r²

Given the following data;

q1 = 2q1

q2 = 2q2

r = r/2

Substituting into the equation, we have;

F = 2q1*2q2/(r/2)²

F = 4q1q2/r²/4

F = 4q1q2 * 4/r²

F = 16q1q2/r²

Therefore, the new force between them is increased by a factor of 16.

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