# Is it possible for opposite charges to repel each other?

1. Sep 27, 2016

### NkaujHelp

1. The problem statement, all variables and given/known data
From rest, how fast would two particles be going at the instant they collide ( if they attract) or when they're infinitely far apart (if they repel)? Explain carefully.

Particle A: 3nC
Particle B: -4nC
Each has a mass of 7 mg. They are .05 m apart.

2. Relevant equations
ΔK + ΔU = 0
ΔK = (1/2)mv(f)^2 - (1/2)mv(i)^2
ΔU = Qa(Vb-Va) <--- I used particle A to find the potential energy at the location of particle B.
V= kQ/r

3. The attempt at a solution
They both start from rest so initial velocity is 0. So the equation I got for the final velocity is SQRT[(kQaQb)/mr]. All I did was added up the kinetic energies of both particles and the electric potential energy of the entire system, which is just those two particles, and set that equal to 0, and solved for final velocity.

So I am assuming that particle B or Qb is a negative charge due to the negative sign, both particles will attract and collide with each other. So the final velocity for both charges the instant they touch or collide with each other is an ERROR. Under that square root, the value is a negative, and you can't really square root a negative.

The reason why I got a negative value under that square root was due to the PE = Q(Vb -Va), where Vb is the electric potential at r=0 (when they touch) and Va at r= .05m apart. Since I can't get a value for the final velocity for attraction, I tried finding the final velocity for when they are really far apart. So in this case, Vb is the electric potential at r = .05 m and Va at r = infinity (when they are far part). By doing this, the negatives cancels out under the square root, giving me a final velocity of 3.8E-6 m/s for when they are really far apart.

Am I overthinking this or did I do something wrong with my algebra here with the equations? Did I set everything up correctly? And if this is on the correct right path, can someone explain to me why I couldn't get a final velocity value for when they collide? Is there any other reasons as to why opposite charges won't attract each other or even have a final velocity at the instant they collide with each other?

Last edited by a moderator: Sep 27, 2016
2. Sep 27, 2016

### andrewkirk

What are you setting to 0 and why? If it is the potential energy then that's not correct. The usual convention for bodies that attract via an inverse square law is to set the potential energy equal to 0 when they are infinitely distantly separated, so that at any finite separation the potential energy is negative.

I don't think this problem is well-posed by the way, because it doesn't tell us the size or shape of the particles. We need to know that to know how far apart their centres of charge will be when they will collide. We presume that the 0.05m distance given is the distance between the centres of charge.

In some problems one assumes particles occupy no space ('point particles') but one cannot do that if one is going to analyse collisions. We need to know at what point they come into contact, which depends on their shape, size and orientation.

Also note that:
• Once the particles get really close to one another their quantum fuzziness starts to dominate and it is impossible to analyse using classical equations, which is what is given above.
• If we had point particles the collision would happen when the separation of the charges was 0, which would make the potential energy infinitely negative, so the energy released by the collision would destroy the entire universe (cue Doctor Who theme music).

3. Sep 27, 2016

### NkaujHelp

I don't understand what you mean by that. I don't get how the separation of the charges = 0 will give a potential energy that is infinitely negative. Can you please explain more. And yes, the particles I was talking about are point particles. My bad for not stating that.

4. Sep 27, 2016

### andrewkirk

What is the value of V when r is 0?

5. Sep 27, 2016

### NkaujHelp

Well, if r = 0, and you divide by a 0, the value of V is undefined. But I still don't understand what's unique about that. I'm confused.

6. Sep 27, 2016

### andrewkirk

What is $\lim_{r\to 0}\frac{kQ}{r}$?

Or, if you haven't covered limits yet, consider what happens when we halve the distance r, initially at 5cm?
A. The absolute value of the PE doubles.
Halve it again and it doubles again.
Halve it again and it doubles again.
As the two particles get closer to one another, the absolute value of the PE increases without limit. Since the PE is always negative, that means it heads towards negative infinity.

7. Sep 27, 2016

### NkaujHelp

Okay. I understand the whole problem now. Thank you so much.

8. Sep 27, 2016

### andrewkirk

You're welcome.