# What happens to the force?

In Coulomb's Law, what happens to the force when the distance between the two charges are doubled? The only thing I'm having trouble is what to put into Q1 and Q2 to find what happens to the force.

gets larger (multiplied by 8)

gets smaller (multiplied by 0.0625)

gets larger (multiplied by 4)

gets larger (multiplied by 2)

gets larger (multiplied by 16)

gets smaller (multiplied by 0.125)

gets smaller (multiplied by 0.5)

can't tell since charges are unknown

gets smaller (multiplied by 0.25)

Relevance

Put in Q1 and Q2.

At original distance d,

F = k * Q1 * Q2 / d²

At double the distance,

F' = k * Q1 * Q2 / (2d)² = ¼ * k * Q1 * Q2 / d² = ¼ * F

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• force drops with square of the distance; doubled distance (2) squared 2^2 = 4, or 1/4 the force or gets smaller multiplied by .25

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• Charge level varies with the inverse square law, like gravity or light intensity.  Multiply by 1/(2^2) or 1/4 or 0.25.

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