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A

Anonymous

Jan 04, 2025

What is the repulsive force between two pith balls that are 13.0 cm apart and have equal charges of -34.0 nC?

What is the magnitude of the repulsive force between two pith balls, each carrying a charge of -34.0 nC, that are separated by a distance of 13.0 cm?

2 Answers

F = (1/4pie) * (q^2/r^2)

= (1/4pie) * (34.0nC)^2/(13.0)^2

= 615 nN

615 nanonewtons is the reulsive force

A
Anonymous

Jan 13, 2025

(9*10^9)(36*10^-9)^2/(.13)^2=6.90*10^-4N or .00069N

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