Electric Field and two point particles

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SUMMARY

The discussion centers on calculating the charges of two point particles separated by 0.4 m with a total charge of 185 µC. When repelling each other with a force of 80 N, the charges are identified as q1=8.05 µC and q2=176 µC, despite their sum not equaling 185 µC. For attraction with the same force, the charges would remain the same in magnitude, but one would be negative. The force calculation utilizes Coulomb's Law, represented as F=k*(|q1*q2|/r²).

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fallen186
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Homework Statement


Two point particles separated by 0.4 m carry a total charge of 185 µC.
(a) If the two particles repel each other with a force of 80 N, what are the charges on each of the two particles?
q1=8.05µC
q2=176 µC
*I know they don't add up to 185 but it took both of the answers as correct*
(b) If the two particles attract each other with a force of 80 N, what are the charges on the two particles?

I don't see why it wouldn't be the same answer except one of them would be negative.
According to F=k*[tex]\frac{|q_{1}*q_{2}|}{r^{2}}[/tex]The type of charge doesn't matter when calculating force.
 
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fallen186 said:

Homework Statement


Two point particles separated by 0.4 m carry a total charge of 185 µC.
(a) If the two particles repel each other with a force of 80 N, what are the charges on each of the two particles?
q1=8.05µC
q2=176 µC
*I know they don't add up to 185 but it took both of the answers as correct*
(b) If the two particles attract each other with a force of 80 N, what are the charges on the two particles?

I don't see why it wouldn't be the same answer except one of them would be negative.
According to F=k*[tex]\frac{|q_{1}*q_{2}|}{r^{2}}[/tex]The type of charge doesn't matter when calculating force.

The product of the signs determine direction, as you observed.
 

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