# Distance of Closest Approach Between Two Charges

1. Jan 18, 2016

### 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?

Last edited: Jan 18, 2016
2. Jan 18, 2016

### 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.

3. Jan 18, 2016

### blue_leaf77

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).

4. Jan 18, 2016

### 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 ?

5. Jan 18, 2016

### gracy

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

6. Jan 18, 2016

### 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.

7. Jan 18, 2016

### gracy

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

8. Jan 18, 2016

### blue_leaf77

$q_1$ 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.
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 $q_2$.

Last edited: Jan 18, 2016
9. Jan 18, 2016

### 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?

10. Jan 18, 2016

### blue_leaf77

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

11. Jan 18, 2016

### gracy

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

12. Jan 18, 2016

### gracy

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

13. Jan 18, 2016

### blue_leaf77

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 $\frac{d\mathbf{r}}{dt} = 0$.
Yes.

14. Jan 18, 2016

### gracy

position vector of charge body which is not moving ?

15. Jan 18, 2016

### gracy

Something seems missing in the above sentence.

16. Jan 18, 2016

### blue_leaf77

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

17. Jan 18, 2016

### gracy

Which two perpendicular vectors?

18. Jan 18, 2016

### blue_leaf77

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

19. Jan 18, 2016

### gracy

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

20. Jan 18, 2016

### 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.

21. Jan 18, 2016

### gracy

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.

22. Jan 18, 2016

### 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.

23. Jan 19, 2016

### gracy

$q_1$ 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?

24. Jan 19, 2016

### cnh1995

Right.

25. Jan 19, 2016

### gracy

Please someone tell me am I right ?