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Two charged particles exert an electric force of 16 n on each other. what will the magnitude of the force be if the distance between the particles is reduced to one-half of the original separation? if the original distance between the particles is now restored, how much larger would the charge on either one of the particles have to become to yield the same force that was present when the particle separation was only one-half its original value?

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wildjay8804
The equation describing the force between two charges (q1 and q2) separated by a distance (d) is .. F = k.q1.q2 / d² 
a) For the original separation .. F ∝ 1/d² .. .. (k, q1, q2 being constant) 
For d/2 .. F ∝ 1/(d/2)² = 4/d² .. 
.. the force has increased by a factor of 4 to (16N x 4) .. ►F = 64 N 

b) To increase the force by a factor of 4 from the original separation .. 

F ∝ q1 x q2 .. .. (k and d being constant) .. 
►increasing either charge by a factor of 4 increases the force 4 times
The equation describing the force between two charges (q1 and q2) separated by a distance (d) is .. F = k.q1.q2 / d²
a) For the original separation .. F ∝ 1/d² .. .. (k, q1, q2 being constant)
For d/2 .. F ∝ 1/(d/2)² = 4/d² ..
.. the force has increased by a factor of 4 to (16N x 4) .. ►F = 64 N

b) To increase the force by a factor of 4 from the original separation ..

F ∝ q1 x q2 .. .. (k and d being constant) ..
►increasing either charge by a factor of 4 increases the force 4 times
Hope this helps! Please mark brainest!

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