Adapted from: Opposites Attract, Don Bahls, University of Alaska - David Newman

Two charges q1 and q2, are separated by a distance d. The magnitude of q1 is 5 × 10 -6C, the magnitude of q2 is 4 × 10 -6C, and the distance between the two charges, d, is 0.5 m. Determine the magnitude of the electric force between q1 and q2.

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Identify known variables: q1 = 5 × 10-6 C, q2 = 4 × 10-6 C, and d = 0.5m

Identify Coulomb constant: kc = 8.99 × 109 N × m2 C2
Equation: FE = kcq1q2 d2 Close Pop Up
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FE = kcq1q2 d2

FE = (8.99 × 109) × (5 × 10-6) × (4 × 10-6) 0.52

FE = 0.1798 0.25

FE = 0.7192 N Close Pop Up

An electric force F exists between two positive charges q1 and q2. If the distance between the two positive charges is increased to 3 times the original distance d, what is the new electric force F' between the two charges? Express your answer in terms of the original electric force F.

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FE is related to the inverse-square of the distance. Close Pop Up
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Original force: F = kc × q1 × q2 d2

New Force: F' = kc × q1 × q2 (3 × d)2 = kc × q1 × q2 32 × d2 = kc × q1 × q2 9 × d2

Pull the 1 9 out in the front of the kc: F' = 1 9 × kc × q1 × q2 d2

Because F = kc × q1 × q2 d2 , you can replace the kc × q1 × q2 d2 in the F' = 1 9 × kc × q1 × q2 d2 with F to get F' = 1 9 × F Close Pop Up