If we take two charges and hold one still, what are the forces?

[annamal]

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 $|F| = 2*\frac{k*q1*q2}{r^2}$

 

[Herman Trivilino]

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]

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.

Since q2 cannot push q1 away, that force does not push q2 away?

If q1 were let go, the total force would include the factor of 2?

 

[Ibix]

Both $q_1$ and $q_2$ experience the same electrostatic force for a given separation whether $q_1$ is fixed in place or not. It's just that in the fixed case, $q_1$ has some other force on it that is equal and opposite to the electrostatic force. This does not affect $q_2$ except in the sense that because $q_1$ does not move the separation between the charges does not grow as rapidly.

 

[Herman Trivilino]

Since q2 cannot push q1 away, that force does not push q2 away?

q2 cannot exert a force on itself. It has nothing to do with q1 being held in place.

If q1 were let go, the total force would include the factor of 2?

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.

 

[annamal]

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.

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?

 

[Ibix]

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?

If they're both free to move then the center of mass does not move - so yes.

 

[Herman Trivilino]

Ok, so the net force of the q1 and q2 charge would be 0 since they exert equal and opposite forces

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]

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).

I mean it is 0 in a system with both q1 and q2 since they cancel.

 

[Herman Trivilino]

I mean it is 0 in a system with both q1 and q2 since they cancel.

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.