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annamal
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If we take two positive charges +q1 and +q2, and we hold q1 still, does q2 experience a repulsive force due to q1 repelling q2 AND a repulsive force due to q2 repelling q1? That is ##|F| = 2*\frac{k*q1*q2}{r^2}##
Since q2 cannot push q1 away, that force does not push q2 away?Mister T said:q2 does not experience a force due to q2.
Whether or not q1 is held in place, the repulsive force exerted on each charged particle is still there. BTW, there's no factor of 2 in the expression for the force.
annamal said:Since q2 cannot push q1 away, that force does not push q2 away?
No, it wouldn't. When q1 is held in place, the net force on q1 is zero. Release q1 and the net force on q1 is ##\frac{kq_1q_2}{r^2}##. The net force on q2 has the same magnitude. The forces act on different objects, so you don't add them together.annamal said:If q1 were let go, the total force would include the factor of 2?
Ok, so the net force of the q1 and q2 charge would be 0 since they exert equal and opposite forces even though there is a net force on q1 and q2 individually?Mister T said:q2 cannot exert a force on itself. It has nothing to do with q1 being held in place.No, it wouldn't. When q1 is held in place, the net force on q1 is zero. Release q1 and the net force on q1 is ##\frac{kq_1q_2}{r^2}##. The net force on q2 has the same magnitude. The forces act on different objects, so you don't add them together.
It's just like me pushing on a block with a force of 10 N. The block pushes on me with a force of 10 N. I don't add them to get 20 N because the forces are acting on different objects.
If they're both free to move then the center of mass does not move - so yes.annamal said:Ok, so the net force of the q1 and q2 charge would be 0 since they exert equal and opposite forces even though there is a net force on q1 and q2 individually?
The net force on q1 is not zero. The net force on q2 is not zero. They exert equal but opposite forces on each other (Newton's Third Law).annamal said:Ok, so the net force of the q1 and q2 charge would be 0 since they exert equal and opposite forces
I mean it is 0 in a system with both q1 and q2 since they cancel.Mister T said:The net force on q1 is not zero. The net force on q2 is not zero. They exert equal but opposite forces on each other (Newton's Third Law).
They do not "cancel". In this context "cancel" means "adds up to zero". But these forces act on different objects. Even if you consider the system consisting of q1 and q2 then the forces are internal. Newton's Third Law requires that forces occur in equal-but-opposite pairs. To say this pair "cancels" is a misconception.annamal said:I mean it is 0 in a system with both q1 and q2 since they cancel.
The formula for calculating the force between two charges is F = k(q1q2)/r^2, where k is the Coulomb's constant, q1 and q2 are the magnitudes of the two charges, and r is the distance between them.
The force between two charges is inversely proportional to the square of the distance between them. This means that as the distance increases, the force decreases and vice versa.
The direction of the force between two charges is always along the line connecting the two charges. If the charges are of the same sign, the force is repulsive, and if they are of opposite signs, the force is attractive.
Yes, the force between two charges can be zero if the charges are of equal magnitude and opposite signs, or if one of the charges is zero. In this case, the charges are in equilibrium and the net force on them is zero.
If one of the charges is held still, the force on the other charge will remain the same. This is because the force between two charges is independent of the motion of either charge.