Electric Field and two point particles

AI Thread Summary
Two point particles separated by 0.4 m have a total charge of 185 µC and repel each other with a force of 80 N, resulting in charges of approximately 8.05 µC and 176 µC, despite not summing to 185 µC. When the particles attract each other with the same force, one charge must be negative, but the absolute values remain the same. The force calculation, using Coulomb's law, does not depend on the sign of the charges, only their magnitudes. The discussion emphasizes that the product of the charges' signs determines the direction of the force. Understanding these principles is crucial for solving problems involving electric fields and forces between charged particles.
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*\frac{|q_{1}*q_{2}|}{r^{2}}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*\frac{|q_{1}*q_{2}|}{r^{2}}The type of charge doesn't matter when calculating force.

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