Distance of Closest Approach Between Two Charges

[gracy]

Come on I was just getting it confirmed from you experts!

 

[jbriggs444]

Come on I was just getting it confirmed from you experts!

Re-read your post 46. It seems more plausible that you are just guessing.

 

[gracy]

Come on I was just getting it confirmed from you experts!

It was for reference point.I was correct there.For Line of action of force I did go through wikipedia but did not understand.

 

[gracy]

=rFsin theta

We will take location of q1 as a reference point.Reasons for this have been discussed in post #36In case of repulsive force applied by q1 on q2 F and r in same direction therefore theta is zero hence no torque is produced.

In case of repulsive force applied by q2 on q1 r i.e position vector will be zero because we have chosen q1 as a reference .Therefore no torque is produced .

And no other force is present to produce torque.Torque is absent that's why angular momentum will be conserved.Right?

 

[jbriggs444]

Google "central force".

 

[SammyS]

@gracy ,
Let's see ...

You made an inquiry regarding using conservation of angular momentum in analyzing the scenario here.

It makes sense to consider the angular momentum of the moving particle with respect to the fixed particle.

To see that this angular momentum is conserved, consider the torque (with respect to the same fixed point) on the moving particle.

The only force on the the moving particle being considered here, is the Coulomb force which the fixed particle (charge, q₁) exerts upon the moving particle (charge, q₂₎).

This force, $\ \vec{F}\,,\$ is along the line determined by the two particles. The position vector, $\ \vec{r}\,,\$ from the fixed particle to the moving particle is along this same line.

What does that imply about $\ \vec{r}\times\vec{F} \ ?\$

 

[gracy]

What does that imply about r⃗ ×F⃗ ?

Theta is zero hence torque is also zero.

 

[gracy]

When line of action of force pass through a point there is no torque (moment of force)produced about that point and that's the reason there is no torque about q1 and q2 .Even if we do not make q1 reference point or even if we will not consider q1 being static we can still conclude that there is no torque when there is system of two charges and only force present is of action -reaction type( as long as the action-reaction forces act along the line connecting the charges. That is, as long as the action-reaction forces are attractive or repulsive. And that's true for electric (Coulomb) forces.)

I hope it is correct.

 

[BvU]

It is.
Also true for graviational force (the attractive part )

Comparable: with a wire you can't pull sideways, only towards you.

 

[SammyS]

You're making some very general statements based on a fairly specific situation. Much care is needed with these statements.

When line of action of force pass through a point there is no torque (moment of force)produced about that point and that's the reason there is no torque about q₁ and q₂ .

(Aren't subscripts so much better to read?)

To be specific: That's the reason that the stationary charge, q₁, exerts no torque (zero torque) on the moving charge, q₂, when that torque is about the location of q₁, the stationary charge.* This is important if you want to use angular momentum conservation to aid in analyzing the trajectory of charge q₂. In this case, the angular momentum of q₂, with respect to the location of charge q₁, is constant.

By the way, using $\ I\omega \$ is not a very helpful way to calculate angular momentum for a single particle unless it has a fixed distance from some point, such as the case of circular motion. The more general way to get the angular momentum about a point (taken here as the origin) is $\ \vec{L} = \vec{r} \times \vec{p}\$.
*To be sure, the torque exerted by q₁ on q₂ is not necessarily zero when it is calculated about some other point.

Note: I hesitated even mentioning this lest the discussion go ambling down some stray path. We're already up to the 60's in the number of posts in this thread.​

 

[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 !