The force of attraction between two like charged table tennis balls is 2.4 * 10-5 newtons. If the charge on the one is 3.8 * 10-8 coulombs and on the other is 3.0 * 10-8 coulombs, what is the distance between the two charges? (k = 9.0 * 109 newton·meter2/coulombs2) A. 0.11 meters B. 0.24 meters C. 0.45 meters D. 0.65 meters

Question

Physics

Carpenter

21 March, 11:19

The force of attraction between two like charged table tennis balls is 2.4 * 10-5 newtons. If the charge on the one is 3.8 * 10-8 coulombs and on the other is 3.0 * 10-8 coulombs, what is the distance between the two charges? (k = 9.0 * 109 newton·meter2/coulombs2) A. 0.11 meters B. 0.24 meters C. 0.45 meters D. 0.65 meters

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Answers (2)

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Guillermo Krause · Answer 1

F = k*Q1*Q2 / r² (Coulombs law)

Now we just put in the values: 2.4*10^{-5} = 9*10^{9}*/frac{ 3.0*10^{-8} * 3.8*10^{-8} }{ r² }/]

Solving for r:

r = squareroot{ 9*10^{9} * / 3.0*10^{-8} * 3.8*10^{-8} }{ 2.5*10^{-5}

Which gives r = 0653835. The answer is D.

Kylee Macdonald · Answer 2

To have a force of 3.8x10-8 the two chaarges must be 0.65m apart.