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

2 Answers

R
Raymond Moen

Feb 20, 2025

F = (1/4pie) * (q^2/r^2)= (1/4pie) * (34.0nC)^2/(13.0)^2= 615 nN 615 nanonewtons is the reulsive force... Show More
R
Raymond Moen

Feb 20, 2025

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

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