# Homework Help: Gauss' Law question

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1. Mar 13, 2016

### Better WOrld

1. The problem statement, all variables and given/known data

Consider two balls of equal radii and masses but opposite charges, distributed uniformly over their volumes. Initially, the balls are at rest and far away from one another. Due to the Coulomb attraction, the balls start moving towards each other. The balls can be treated as charged clouds, that is to say, they can interpenetrate without friction. During the interaction, the maximum speed achieved by the balls is $v_{max}=10~\text{m}/\text{s}$. What is the magnitude of their maximum acceleration in meters per second squared? The radius of the balls is $R=1~\text{m}$.

2. Relevant equations

$$\int E.da=q/\epsilon$$

3. The attempt at a solution

A friend gave me this question and told me that it could be solved using Gauss' Law. However, I don't understand the question properly. What is the significance of 'balls are at rest and far away from one another'? Also, what does 'balls can be treated as charged clouds, that is to say, they can interpenetrate without friction' mean? Lastly assuming that the balls act only under the influence of Electrostatic force of attraction $F=k\dfrac{q_1q_2}{r^2}$ shouldn't acceleration be maximum when the balls are about to collide?

I would be really grateful if somebody would kindly show me how to solve this problem. Many thanks!

2. Mar 13, 2016

### haruspex

It allows you to work out the initial energy.
But they do not collide... they pass through each other.
First step is to figure out when their speeds will be max.

3. Mar 13, 2016

### Better WOrld

So initially, the total energy would be 0 (assuming U - Potential Energy at $\infty$ is 0).
Shouldn't the speed be maximum at the instant they pass through each other? Also by conserving energy, at the moment the balls pass through each other, r (separation between them) becomes 0. Thus U would become $-\infty$. Thus $v$ would become $infty$.
And how would I proceed from this point?

4. Mar 13, 2016

### haruspex

Passing through each is not an instantaneous event. But yes, it will be a max when their centres coincide.
No, quite finite. These are not point charges. At the instant the spheres' centres coincide, how would you describe the charge distribution?

5. Mar 13, 2016

### Better WOrld

I'm not really sure about this, but the balls would be similar to solid charged spheres. Now the field (and consequently the force due to the sphere) at any external point is as though the sphere acts as a point particle. So in this case, why should the Potential Energy not be $-\infty$?

Also I can't understand how the charge distribution would bee. I still can't understand the meaning of '

6. Mar 13, 2016

### haruspex

Why solid spheres? The charge is only on the surface. Once they start to pass into each other, you can no longer use the external point view.
You still seem to be struggling with the interpenetration concept. It is tricky because you could not physically construct something like this. You have to pretend that each spherical shell, though quite rigid in itself, can pass through the other. The only interaction between the shells is from their charge clouds.
So, consider the instant at which the centres coincide. Since the spheres have the same radius, their shells coincide exactly. If each has a surface charge density ρ, how would you describe the charge distribution at that instant?

7. Mar 14, 2016

### Better WOrld

Sir, assuming that I understand properly, would the surface charge density become $2\rho$?

8. Mar 14, 2016

### haruspex

Yes. Can you figure out the potential energy in that state?

Edit: that was wrong.... see below.

Last edited: Mar 14, 2016
9. Mar 14, 2016

### Better WOrld

Could you please show me how to calculate it? I know that the $Potential$ at its center is $-\dfrac{kQ}{R}$

10. Mar 14, 2016

### haruspex

If that's the potential at the centre, what is the potential at the surface?

11. Mar 14, 2016

### Better WOrld

Isn't it the same?

12. Mar 14, 2016

### haruspex

No, sorry... I forgot the problem statement. These two spheres have opposite charges.

13. Mar 14, 2016

### Better WOrld

So the charge would get neutralised ie there would be no charge on the surface?

14. Mar 14, 2016

### haruspex

Yes.
So we need a way of assessing the total PE change as the two spheres go from being a long way apart to having merged and neutralised each other. One way is to consider this happening in small packages of charge instead of the whole sphere moving together.
Suppose at some stage there remains a charge of -q on one and +q on the other. A small charge dq now moves from the +q to the -q. What is the change of PE in the system?

15. Mar 14, 2016

### Better WOrld

Sir I'm not yet comfortable with the idea of small packages of charge. How would 'packages of charge' move in place of the spheres themselves. If the spheres themselves are not moving as individual units, why is it mentioned in the question that the $balls$ can interpenetrate?
Thus, I'm not able to calculate the change in PE. I know that $d(PE)=-dW$. However, $W=\int F.dr$ But what is $F$? Shouldn't $dq$ be repelled by the quantity of charge $+q-dq$ and attracted by the charge $-q$? Then wouldn't $F$ be the vector sum of these 2 forces? Also what would the limits on the Integral be?

16. Mar 14, 2016

### haruspex

We can use a rather easier integral, $\int V.dq$. You can write V in terms of q, the charge magnitudes at that instant, and R. But yes, V is the potential difference between the two, so there is a doubling up.
What do you think the range for q is?
(I feel there must be a way of doing this without integrals, but I cannot think of one.)

17. Mar 14, 2016

### Better WOrld

Sir please could you show me how to write V in terms of q, the charge magnitudes at that instant and R? Could you also explain what you mean by 'there is a doubling up'?
Sir, $Q\to 0$ since charge packets move gradually away from the sphere with $+Q$ charge initially.
Lastly Sir, I think that somehow we are supposed to use Gauss' Law (this question, so I am told, was actually a part of practice problems on Gauss' Law).

18. Mar 14, 2016

### haruspex

When there is a charge of q on one sphere, what is its potential? You already answered this. The other sphere has charge -q at this time, so what is its potential? What is the potential difference between the two?

19. Mar 14, 2016

### Better WOrld

Sir, is it $-2k\dfrac{kq}{r}$ where $r$ is the separation between the centers of the 2 spheres?

20. Mar 14, 2016

### haruspex

No, the spheres are a long way apart here, so they do not influence each other. Each has a potential in its own right, nothing to do with the distance between them.

21. Mar 14, 2016

### Better WOrld

Sir isn't potential with respect to a point? So which point do we take potential about?

The potential of any spherical shell at an interior point is $-kQ/R$ where R is its radius.

22. Mar 14, 2016

### haruspex

Yes, and as we noted that is the potential at the surface too. So how much work do we have to do to take a charge dq from it away to infinity (where the potential due to that sphere is zero).

23. Mar 16, 2016

### Better WOrld

Sir I tried, but I couldn't calculate it. Please could you show me how to?

24. Mar 16, 2016

### ehild

From the OP:
Far away from each other, the potential energy of the system of the two balls is -kQ2/d. When they just touch each other, d=2R. Assume a Gaussian surface that encloses the two balls. Keep this surface. When interpenetrating, the -ve and +ve charges neutralize each other in the common volume . What happens with the electric field around the spheres? What is the net charge and the electric field when the centres coincide? What is the electric field on the Gaussian surface? What is the electric energy then?

25. Mar 16, 2016

### haruspex

When the charge on the sphere is q, the potential at the sphere is -kq/R. The potential at infinity is zero.
If we move a small charge dq from the sphere to infinity, what is the increase in the potential for that charge? How much work does it take?