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

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

chillout · Answer

$$F = k*\frac{ Q1*Q2 }{ r² }$$  (Coulomb's 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:
$$r = \sqrt{ 9*10^{9}*\frac{ 3.0*10^{-8} * 3.8*10^{-8} }{ 2.5*10^{-5} }}$$ Which gives r = 0653835. The answer is D.