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How far apart must two point charges of 75.0 $\mathrm{nC}$ (typical of static electricity be to have a force of 1.00 $\mathrm{N}$ between them?
7.1 $\mathrm{mm}$
Physics 102 Electricity and Magnetism
Chapter 18
Electric Charge and Electric Field
University of Washington
University of Winnipeg
McMaster University
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force between two charges is given by when divided by full, perhaps alone, not Q. One Q two, divided by distant square queue and queue to being two charges. Now we are given their forces when Newton, for one divided by four perps are not has a value nine times to empower. Nine. Both charges have same value that is 75 Nanako. So it's just squared, divided by distance where sinister square is simply nine times 10 power 75 square times 10 power *** to night. So nine times 75 square is 50,000 6 25 Simplifying that will give me 5.6 to 5 times 10 Power negative fight. So this also can we return eyes 50.6 to 5 times 10 Power negative six Because then taking square root will be easier. 15.6 to 5 has a square root 7.11 square root of 10 power negative sixes 10 power negative three Just a man you are so answer in this case should be just two significant digits of 7.1 millimeters. I hope you understand
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