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The force of repulsion between two like–charged table tennis balls is 8.2 × 10-7 newtons. If the charge on the two objects is 6.7 × 10-9 coulombs each, what is the distance between the two charges?

(k = 9.0 × 109 newton·meters2/coulomb2)

A.
0.32 meters
B.
0.70 meters
C.
6.7 meters
D.
8.2 meters
the -7, -9, and 9 are exponets

1 Answer

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Using Coulomb's law, the distance between the two like-charged table tennis balls with a force of repulsion of 8.2 x 10^-7 newtons and charges of 6.7 x 10^-9 coulombs each is found to be 0.70 meters. Here option B is correct.

To calculate the distance between the two charges using Coulomb's law, we apply the formula:

F = k * (|q_1 * q-2|) / r^2

Where F is the force between the charges, k is Coulomb's constant (9.0 × 10^9 N·m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.

We know the force of repulsion (F) is 8.2 × 10^-7 newtons and each charge (q_1 and q_2) is 6.7 × 10^-9 coulombs. We can now rearrange the formula to solve for the distance r:

r^2 = k * (|q1 * q2|) / F

Plugging in the values:

r^2 = (9.0 × 10^9 * (6.7 × 10^-9 * 6.7 × 10^-9)) / 8.2 × 10^-7

r^2 = (9.0 × 10^9 * 44.89 × 10^-18) / 8.2 × 10^-7

r^2 = (404.01 × 10^-9) / 8.2 × 10^-7

r^2 = 0.4927 × 10^-2

Then, r = √(0.4927 × 10^-2)

r = √4.927 × 10^-4

r = 0.70 meters

Therefore, the distance between the two charges is 0.70 meters. Here option B is correct.

User Kemin Zhou
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