two charges separated by 60cm are moves far a part untill the force between them has decreased by a factor of 4. How a part are the changes them ?
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two charges seprarated by a 60 cm which means 0.6m
means r = 0.6m
f1=q1q2/r1^2
now force is decreased by a factor of 4
so f1= f1-4
(f1-4)=kq1q2/r2^2
divide the equation
f1/(f1-4) = r2^2/0.6
r2^2= (f1/(f1-4) 0.6^2
r2= √ (f1/(f1-4) 0.6^2 = 0.6√(f1/(f1-4)
means they will apart from each other with a distance of 0.6√(f1/(f1-4)
means r = 0.6m
f1=q1q2/r1^2
now force is decreased by a factor of 4
so f1= f1-4
(f1-4)=kq1q2/r2^2
divide the equation
f1/(f1-4) = r2^2/0.6
r2^2= (f1/(f1-4) 0.6^2
r2= √ (f1/(f1-4) 0.6^2 = 0.6√(f1/(f1-4)
means they will apart from each other with a distance of 0.6√(f1/(f1-4)
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