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## At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is is decreased by a

Question

At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is is decreased by a factor of 3 and the charge on one of the balloons is doubled, then the repulsive force would be?

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Physics
6 months
2021-07-26T08:56:04+00:00
2021-07-26T08:56:04+00:00 1 Answers
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## Answers ( )

Answer:

2.9 N

Explanation:

When the separation distance, r, is 0.5 m, the electrostatic force is 0.32 N. Electrostatic force is given as:

F = (k * q1 * q2) / r²

Where F = force acting on the balloons

k = Coulombs constant

Therefore:

0.32 = (k * q1 * q2) / 0.5²

=> k * q1 * q2 = 0.32 * 0.5² ————(1)

When the distance is decreased by 3, that is r = r/3 = 0.5/3

F = (k * q1 * q2) / (0.5/3)² ————(2)

Putting (1) into (2):

=> F = (0.32 * 0.5²) / (0.5/3)²

F = (0.32 * 0.5² * 3²) / 0.5²

F = 2.9 N

Therefore, the force would be 2.9 N