Distance of Closest Approach Between Two Charges

[jbriggs444]

The whole paragraph is as follows

We consider a charge $q_1$ which is again fixed, and always kept at rest and from this charge particle .of closest approach. here, we consider a charge $q_1$ which is again fixed, and always kept at rest and from this charge particle $q_1$ at an impact parameter d another charge which is having a mass m and charge $q_2$ is thrown. with the speed $v_0$. We know that $q_2$ as the charge gets closer to this $q_1$. due to repulsion its path will be deviated .We wish to find out the distanceof closest approach of the 2 particles. that will be at a position when the velocity vector of. the charge $q_2$ will be at right angle to the line joining $q_1$and $q_2$ that will be the distance of closest approach we can write as $r_0$. and in this situation we know as this $q_1$ is always fixed and the line of action of force will always pass through .$q_1$ we can use angular momentum conservation as.the electric force won't apply any torque with respect to the charge i$q_1$ in the whole system.


is $q_1$ a reference point ?or location of $q_1$ is reference point ?Can we use any point as a reference point?That is in this case can I take $q_2$ as a reference point.I think No because $q_2$ is moving it's location keeps on changing.

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.

 

[nasu]

Torque is the tendency of a force to rotate an object about an axis.

So what doe this tells you about the torque when r and F are along the same direction?

 

[gracy]

So what doe this tells you about the torque when r and F are along the same direction?

Torque=rFsin theta
theta will be zero hence torque will also be zero.

 

[gracy]

I did not understand properly what purpose does reference point serve in torque?

 

[jbriggs444]

I did not understand properly what purpose does reference point serve in torque?

If you do not have a reference point, how can you compute the r in

 

[nasu]

How do you define that r if you don't have a reference point? Any position vector is relative to some origin.
The torque will depend on what point do you choose for origin. It may be zero for one particular origin and non-zero for another origin.

 

[SammyS]

The whole paragraph is as follows

"We consider a charge q₁ which is again fixed, and always kept at rest and from this charge particle .of closest approach. here, we consider a charge $q_1$ which is again fixed, and always kept at rest and from this charge particle $q_1$ at an impact parameter d another charge which is having a mass m and charge $q_2$ is thrown. with the speed $v_0$. We know that $q_2$ as the charge gets closer to this $q_1$, due to repulsion its path will be deviated .We wish to find out the distance of closest approach of the 2 particles. That will be at a position when the velocity vector of the charge $q_2$ will be at right angle to the line joining $q_1$and $q_2$. That will be the distance of closest approach we can write as $r_0$. And in this situation we know as this $q_1$ is always fixed and the line of action of force will always pass through $q_1$. We can use angular momentum conservation as the electric force won't apply any torque with respect to the charge i$q_1$ in the whole system."

I believe that you have some 'cut and paste' error in the above. I have done a 'strike-through' through the repeated text and capitalized a few words I was confident started sentences as well as removing some random periods .

Is this essentially the question you began in your OP and then expanded in post #4?

Does this paragraph come from some particular source ?

 

[gracy]

. This follows because of the way the vector cross product works -- two identical forces applied along the same line of action will result in identical torques.

I thought the two lines of action of repulsive forces are different(opposite direction).​

 

[jbriggs444]

I thought the two lines of action of repulsive forces are different(opposite direction).​

Yes, if two forces are identical (same direction) and on the same line of action, they yield identical torques. If two forces are equal but opposite (opposite directions) and are on the same line of action, they yield equal but opposite torques.

 

[gracy]

What is line of action of force in below picture

Is it r?

 

[gracy]

What is line of action of force in below picture

 

[jbriggs444]

The "line of action" of a force is a line that goes through the point of application. It extends forward in the direction of the applied force and backward in the opposite direction. For a force applied at a distance, the line of action will go through both the source of the force and its point of application. The two forces in a 3rd law force pair will always share the same line of action.

In the drawing you show, the line of action of force F is hiding under the vector F. The artist specifically placed F where he did in order to make its point of application (and, thus, its line of action) obvious.

If you had Googled the term, you could have learned this easily. https://en.wikipedia.org/wiki/Line_of_action

 

[gracy]

Is vertex of angle θ reference point in the following

 

[Vanadium 50]

If you had Googled the term, you could have learned this easily. https://en.wikipedia.org/wiki/Line_of_action

That is true. She could have. But she didn't. And it turned out she didn't need to.

I wrote this a month ago. It's still true.

Gracy, I don't think PhysicsForums is helping you. You've asked us about vectors in the past, and now it's clear you haven't learned them.

The problem is that you immediately jump to asking a question here without having put much work into it, and when you are guided by someone towards the answer, you don't try and work it out for yourself, but instead ask for another hint. And another. And another. Eventually, you have been hinted all the way to the answer. Well, you've gotten the answer, but you haven't really learned.

You're going to have to decide if you want to learn or not. If you want to learn, you are going to have to spend more time thinking and working on your own. In this case, your starting point for basic vectors shouldn't be asking for more hints - it should be going back to your past materials and see if you can solve this using what you have already been told.

You can see it here -she's on E&M but is struggling with mechanics. Why is she struggling with mechanics? Because she didn't do the work when she was taking it - she got us to do it for her. Is she learning E&M? Nope - same reason. And it's our fault.

 

[jbriggs444]

Is vertex of angle θ reference point in the following

Yes.

It is common practice in high school physics is to talk about angular momentum in two dimensions only and to use the notion of an "axis of rotation" rather than a point of reference.

 

[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