Distance of Closest Approach Between Two Charges

[gracy]

they yield equal but opposite torques

Which results in zero torque.

 

[jbriggs444]

Which results in zero torque.

That depends on what you are talking about.

If you are talking about net torque on the system within which both objects are members, then the net torque on that system is zero, yes. That is one way of coming up with the idea that angular momentum must be conserved in a closed system.

If you are talking about the torque on one object or the other then the torque on that object will not be zero [except for the corner case when the "equal but opposite" torques are both zero].

 

[gracy]

corner case

What's that?

 

[jbriggs444]

What's that?

You really do need to use google. Or read books.

 

[gracy]

You really do need to use google.

Shall I type" corner case in torque"?

 

[gracy]

corner case

I think you meant extreme case and that's what I asked when and how that occurs in case of torque ?

If you are talking about the torque on one object or the other then the torque on that object will not necessarily be zero

I think I have got enough and helpful answers in this thread .Thanks .Not going to ask any further questions on this.

 

[SammyS]

The location of $q_1$ is a good choice for a reference point. The fact that it is not moving (in your chosen coordinates) is one reason. Another good reason is because whatever external force holds $q_1$ in place exerts no torque if we choose the location of $q_1$ as the reference point.

In my opinion:

The fact that q₁ is stationary, implies that the system of the two particles (charges) is not closed (isolated). There are external forces on q₁, keeping it stationary as it interacts with q₂. There is a non-zero torque on this system about its center of mass.​

That being said, you can consider the two charges as a system, and as jbriggs444 states, " whatever external force holds $q_1$ in place exerts no torque" {on the system} "if we choose the location of $q_1$ as the reference point."

 

[SammyS]

In a pm, gracy asked:

Why are we even taking angular momentum into account{?} I can't see any rotational motion.I mean what will I take in place of r in angular momentum formula L=mvr as r has to be radius of rotation.
​

See Posts #28 and #35. They're gracy's posts.

See what briggs said in post # 34.

Calculate angular momentum as I suggested in post #60.
(I've got to get to my office now. Let briggs continue to help, as he is able.)

(This post has been edited slightly. No relevant content was changed.)

 

[jbriggs444]

The gentle advice from Vanadium 50 in #49 seems most appropriate. All that need be said has been said. We are doing gracy no favors by responding.

 

[gracy]

Yes I am done with this thread

 

[BvU]

Same here !