What is the repulsive force between two pith balls that are 4.00 cm apart and have equal charges of - 20.0 nC? k 8.988 x10^9 (N-m^2)/C^2

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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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What is the repulsive force between two pith balls that are 4.00 cm apart and have equal charges of - 20.0 nC?
k =8.988 x10^9 (N-m^2)/C^2
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Transcribed Image Text:What is the repulsive force between two pith balls that are 4.00 cm apart and have equal charges of - 20.0 nC? k =8.988 x10^9 (N-m^2)/C^2 Edit View Insert Format Tools Table 12pt v Paragraph v |BIUA v 总|
Expert Solution
Step 1

Given data:

Distance between the charges (d) = 4 cm

Charge on each ball (q) = -20μC

k = 8.988*109 N.m2/C2

Need to determine the repulsive force between the charges.

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