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lyudmila [28]
3 years ago
8

Two charged objects attract each other with a force of F. What happens to the force between them if one charge is doubled, the o

ther charge is tripled, and the separation distance between their centers is reduced to ¼ its original value?
Physics
1 answer:
Ray Of Light [21]3 years ago
3 0

Answer:

The force is increased by a factor of 96. (F1 = 96F)

Explanation:

Let the first charge be Q1

Let the second charge be Q2

Let the distance between their centers be r

The electrostatic force, F, between them is:

F = (k*Q1*Q2) / r²

If the first charge is doubled = 2Q1

If the second charge is tripled = 3Q2

The separation between their centers is reduced to ¼ = ¼r

The force between them becomes:

F1 = (k* 2Q1 * 3Q2) / (¼r)²

F1 = (6 * k * Q1 * Q2) / (r²/16)

F1 = 96(k * Q1 * Q2) /r²

F1 = 96F

The force is increased by a factor of 96.

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This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

V_{f}=V_{o}+at (1)  

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V_{f}=0+at (3)  

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