Distance of Closest Approach Between Two Charges

[gracy]

There is charge $q_0$ which is fixed.from a distance far away another charge $q_1$ is fired towards it with speed $V_0$ then we know as both of these are positive charges these will repel each other as the charge $q_1$ comes closer. While coming closer to $q_0$, it's speed decreases and after reaching a closest distance $r_0$ it's speed becomes zero . Now this is the position when , both of these charges will be at rest and there won't be any kinetic energy in $q_1$ left at this position.
My first question is how charge $q_0$ must have been fixed such that it does not move due to repulsion?

 

[cnh1995]

I'm not sure, but one way to do this would be putting the charge q₀ on a spherical shell (non conducting) such that it uniformly distributes over it. This way, the charge q₀ will be fixed on the shell as q1 approaches it.

 

[blue_leaf77]

My first question is how charge q0q0q_0 must have been fixed such that it does not move due to repulsion?

The situation mentioned in your problem description is the instant moment when the charge $q_1$ is at rest during its course of motion. Of course as you may have predicted, after this infinitesimally short moment of stoppage, $q_1$ will be experiencing repulsion and will thence move back to where it came from (assuming the impact parameter is 0).

 

[gracy]

Ok.We consider a charge $q_1$ which is again fixed, and always kept at rest. and from this charge particle at an impact parameter d another charge, which is having mass m and charge $q_2$ is thrown with the speed $V_0$. We know that as the charge $q_2$ gets closer to. $q_1$ due to repulsion it's path will be deviated .We wish to find out the distance of closest approach of the 2 particles.I have read that
distance of closest approach 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 $q_2$
I want to know why ?

 

[gracy]

q1q1q_1 will be experiencing repulsion and will thence move back to where it came from (assuming the impact parameter is 0).

But you are talking about $q_1$ and I had asked about $q_0$ .

 

[blue_leaf77]

Ah my bad, I misunderstood your question. I thought you were questioning how charge $q_1$ can be fixed all the time after it reached the closest distance with $q_2$. It turns out that your question is about the way to hold $q_0$ fixed. Well to answer this, just imagine $q_0$ is held by something, like hypothetical human's hand.

 

[gracy]

I would also like to understand /know why $q_1$ stops?Does repulsive force balance the force with which $q_1$ was moving?

 

[blue_leaf77]

I would also like to understand /know why $q_1$ stops?Does repulsive force balance the force with which $q_1$ was moving?

$q_1$ can stop during its motion only when the impact parameter is 0. You can understand this from the conservation of energy, $T(x)+U(x) = C$ where $C$ is constant. Charge $q_1$ started from infinity with certain initial velocity, also initially $U(-\infty) = 0$. As $q_1$ approaches the fixed charge at the origin, $U(x)$ will be increasing, on the other hand $T(x)$ will decrease in order to keep the energy conservation satisfied. At some point $T(x)$ will equal 0, that's the moment when $q_1$ stops.

distance of closest approach will be at a position when the velocity vector of. the charge q2q2q_2 will be at right angle to the line joining q1q1q_1and q2q2q_2
I want to know why ?

Solving the equation of motion, you will see that elastic collision due to Coulomb repulsion yields a hyperbolic trajectory of the moving body. You should be able to see intuitively that the closest distance occurs when the velocity vector is perpendicular to the position vector of $q_2$.

 

[gracy]

The only difference between the situations I mentioned in OP and in my post #4 is impact parameter.In OP it is zero .Right?

 

[blue_leaf77]

The only difference between the situations I mentioned in OP and in my post #4 is impact parameter.In OP it is zero .Right?

If in your OP the impact parameter is 0, then yes there will certain instant when $q_1$ is at rest.

 

[gracy]

distance of closest approach will be at a position when the velocity vector of. the charge q2q2q_2 will be at right angle to the line joining q1q1q_1and q2

Does this happen only when impact parameter has some non zero value?

 

[gracy]

In post #8 by $T(x)$ did you mean kinetic energy ?

 

[blue_leaf77]

Does this happen only when impact parameter has some non zero value?

In general, you can say that when the closest distance is reached, the vector relation
$$
\mathbf{r} \cdot \frac{d\mathbf{r}}{dt} = 0
$$
,with $\mathbf{r}$ the position vector, holds. When the impact parameter is different from zero, that equation is satisfied because the of the dot product between two perpendicular vectors. When the impact parameter is zero, that equation is satisfied because $\frac{d\mathbf{r}}{dt} = 0$.

In post #8 by $T(x)$ did you mean kinetic energy ?

Yes.

 

[gracy]

with r the position vector

position vector of charge body which is not moving ?

 

[gracy]

that equation is satisfied because the of the dot product between two perpendicular vectors.

Something seems missing in the above sentence.

 

[blue_leaf77]

position vector of charge body which is not moving ?

In general $\mathbf{r}$ is the relative vector between the fixed charge and the moving charge. If the fixed one is held at the origin, this vector becomes the position vector of the moving charge.

Something seems missing in the above sentence.

It should be "that equation is satisfied because of the dot product between two perpendicular vectors."

 

[gracy]

between two perpendicular vectors.

Which two perpendicular vectors?

 

[blue_leaf77]

$\mathbf{r}$ and $\frac{d\mathbf{r}}{dt}$.

 

[gracy]

You mean rate of change of position vector i.e $\frac{dr}{dt}$ is also a vector?

 

[blue_leaf77]

Yes, it's also a vector. Because the change of a vector may involve the change of its direction and this change of direction must be accounted for in its rate of change.

 

[gracy]

When the impact parameter is zero, that equation is satisfied because dr/dt = 0.

Did you mean when the impact parameter is zero,rate of change of position vector is also zero?But I think magnitude of rate of change of position vector will change.

 

[nasu]

The rate of change of the position vector (which is the velocity, by the way) will be zero at the point of minimum distance only. This is, for impact parameter equal to zero.
For this case, the velocity is initially directed towards the fixed charge. Its magnitude decreases until it reaches zero at the minimum distance and then start increasing, but in the opposite direction. For this case the velocity is all radial, there is no component perpendicular to the line connecting the two charges.

For non-zero impact parameter, the radial component of the velocity have a similar behavior. But there is a normal component as well. This component does not vanish. The electrostatic force is always radial so there is no normal component of the force.
The dot product $\vec{r} \cdot \frac{d \vec{r}}{dt} = \vec{r} \cdot \vec{v}$ is zero when the velocity has no radial component, so the mobile charge it is neither approaching nor moving away form the fixed charge. But it still have some normal component.

By "normal" in the above I mean perpendicular to the radial direction, with center in the fixed charge.

 

[gracy]

Thank you so much the above answer is really helpful.

$q_1$ is always fixed and the line of action of force will always pass through $q_1$.

I did not understand this.

I know the line of force is a geometric straight line in the direction of the direction of the force and through the point at which the force is being applied.

Here the force is force of repulsion and the line of action of this repulsive force will always pass through $q_1$ Right?

 

[cnh1995]

I know the line of force is a geometric straight line in the direction of the direction of the force and through the point at which the force is being applied.

Right.

 

[gracy]

Thank you so much the above answer is really helpful.

q1q1q_1 is always fixed and the line of action of force will always pass through q1q1q_1.

Here the force is force of repulsion and the line of action of this repulsive force will always pass through q1q1q_1 Right?

Please someone tell me am I right ?

 

[nasu]

Well, what does Coulomb's law tells you about this? What is the direction of the electrostatic force?
https://en.wikipedia.org/wiki/Coulomb's_law
Why do you even have to ask?

And what is "q1q1q_1"?

 

[BvU]

nasu: q1q1q_1 is what you get when you quote $q_1$.

Please someone tell me am I right ?

yes, the electrostatic force between two charges is collinear with the vector joining the charges.

 

[gracy]

And then

We can use angular momentum conservation as the electric force won't apply any torque with respect to the charge $q_1$ in the whole system

Please help me to understand this line.

 

[nasu]

Do you know the definition of angular momentum and in what conditions will be conserved?

 

[gracy]

Do you know the definition of angular momentum

It is the product of its moment of inertia and its angular velocity.

and in what conditions will be conserved

When there is no torque.

 

[nasu]

OK, and how do you calculate torque? What is the definition?

 

[gracy]

OK, and how do you calculate torque? What is the definition?

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

 

[gracy]

My question is

electric force won't apply any torque with respect to the charge q_1 in the whole system

Is it the same thing as electric force won't apply any torque on the charge q_1 in the whole system.

 

[jbriggs444]

Is it the same thing as electric force won't apply any torque on the charge q_1 in the whole system.

Angular momentum is only given by moment of inertia multiplied by angular rotation rate if we are talking about a rigid object that is rotating. We are not talking about a rotating rigid object in this case. So we have to use the more general definition for angular momentum.

The more general definition for angular momentum requires a reference point. The angular momentum of a system is the sum over all the particles in the system of the vector cross product of the momentum of the particle times its displacement from the reference point.

Taking angular momentum "with respect to" a point means using that point as the reference.

Edit: The electric force applies no net torque to the entire system because any torque applied by $q_1$ on $q_2$ will be equal and opposite to the torque applied by $q_2$ on $q_1$. 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.

 

[gracy]

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.

 

[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.