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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

Verified Answer :

baedenc

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

johnkellock

F = kQ1Q2 / r²  (Coulombs law)

Now we just put in the values: 2.410^{-5} = 910^{9}\frac{ 3.010^{-8} 3.810^{-8} }{ r² }\]

Solving for r:

r = squareroot{ 910^{9}/ 3.010^{-8} 3.810^{-8} }{ 2.510^{-5}

Which gives r = 0653835. The answer is D.

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