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.