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-

Question

4 years ago

10

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

Physics

2 answers:

k0ka [10] · Answer 1 · 2022-02-21T14:05:19+03:00

k0ka [10]4 years ago

7 0

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

Radda [10] · Answer 2 · 2022-02-21T15:12:30+03:00

Radda [10]4 years ago

3 0

<span>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.</span>