Coulomb's Law and repulsive force

AI Thread Summary
The discussion centers on applying Coulomb's Law to determine the effect of increased separation on the repulsive force between two charges. When the separation is increased to 160% of the initial distance, the new force is calculated to be approximately 5.86 N, which aligns closely with the provided option of 5.9 N. Additionally, when the distance between two charged bodies is doubled, the force reduces to one-fourth of its original value, confirming the inverse square relationship. The calculations demonstrate a solid understanding of the principles involved. Overall, the thread effectively illustrates the application of Coulomb's Law in varying scenarios.
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Homework Statement



Charges q1 and q2 exert repulsive forces of 15 N on each other. What is the repulsive force when their separation is increased so that their final separation is 160% of their initial separation?

a.1.3 N

b.9.8 N

c.8.7 N

d.5.9 N


Homework Equations



F = (k*q_1*q_2)/(r^2)

The Attempt at a Solution



F_2 = [k*q_1*q_2]/[(1.6^2)*(r^2)]

F_2 = F/(1.6^2) = F/2.56

F_2 = 15 N/2.56 = 5.86 N ??




Homework Statement



The force between two very small charged bodies is found to be F. If the distance between them is doubled without altering their charges, the force between them becomes


a.F/2

b.2F

c.F/4

d.4F


Homework Equations



F = (k*q_1*q_2)/(r^2)

The Attempt at a Solution



F_2 = (k*q_1*q_2)/(2r)^2 = (k*q_1*q_2)/(4*r^2)

F_2 = F/4 ?


Thanks.
 
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Yeah I think you've nailed the inverse square concept.
 
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