# Flux through far away disc and potential Q.

1. Feb 2, 2013

### CAF123

1. The problem statement, all variables and given/known data
1) A small disc of radius R with surface normal $\hat{k}$ is placed a distance z from a point charge −q. Assuming that R ≪ z, derive an expression for the electric flux $\Phi_E$ passing through the disc.

2) Twenty-seven drops of salt water with the same radius are each charged to a potential of V volts. They are made to coalesce into a single drop. How does the potential of the new single drop compare with that of the original drops?

3. The attempt at a solution

1)If R << z, then the disc will look like a point charge. So the E field a distance z away will be $\vec{E} = -\frac{-q}{4\pi \epsilon_o z^2}\hat{k} = \frac{q}{4 \pi \epsilon_o z^2}\hat{k}$. (This result does not make sense - the E-field should be pointing in negative z, but the negative cancels?) EDIT: I see that the E-field points in -ve z because the charge is negative, so I have essentially counted the same thing twice.
If I say that the disc is like a point charge, then how would I find the flux? For a disc, A = πR2, but I am unsure about how to apply this to the point charge.

2) I think this is simply 27V,since the potentials can be added like scalars (scalar summation). I don't know what else to say here. If the potential across the drop is V then one side of the drop is higher in potential than the opposite side by V volts? Then since potentials are scalar additive, I just add all the potential differences together and get 27V? Is that right? Maybe I should also say that the work required to put these drops together is W = 27(qV), q the charge on a single drop.

Last edited: Feb 2, 2013
2. Feb 2, 2013

### TSny

1) For Z >> R the E-field will essentially be uniform over the disk and perpendicular to the disk.

2) How does the potential of a charged sphere depend on the charge and radius? What would be the charge and radius of the final sphere (compared to the initial drops)?

3. Feb 2, 2013

### CAF123

If z >>R, would the disk not look like a point charge? Is what you said above not apply to the case z <<R?
The potential of a charged sphere is an equipotential: V = $\frac{Q}{4 \pi \epsilon_o R}$. I can imagine if the drops are all clumped together, the radius would increase by 27 and if we are adding charge as well that would be 27 x as much charge as well. So, $V = \frac{27Q k}{27 R} = \frac{Q k}{R},$ the same as that of the original drop?

Last edited: Feb 2, 2013
4. Feb 2, 2013

### Epistimi

Here is what I do not understand:

If you claim the radius of the disc is negligible in comparison to the distance z away from the point charge, the expression for flux seems to go to zero? Does it not?

Last edited: Feb 2, 2013
5. Feb 2, 2013

### TSny

In the limit of z going to infinity, the flux through the disk would go to zero. But, you don't want to go to that extreme limit. You want a non-zero approximation for the flux when z>>R.

An analogy would be the sun's rays hitting a piece of paper held perpendicularly to a radial line from the sun. In the limit of carrying the paper infinitely far away from the sun, the flux of light hitting the paper would go to zero. But, that's too trivial.

What if the paper is here on the earth? Then the flux of light is not zero. But the earth is far enough away from the sun and the area of the paper is small enough that you can consider all the rays of light that strike the paper to be parallel to the normal to the surface of the paper. Also, the intensity of the light may be considered uniform over the surface of the paper.

The radius would not increase by a factor of 27. How would the volume of the final sphere compare to the volume of one of the initial drops?

6. Feb 3, 2013

### CAF123

I see. I was taking the extreme limit before, that's why I didn't understand how to compute a non zero flux, but It makes sense now.

So I am integrating over the whole disk, with the E field perpendicular to the surface normals across the disk. This means: $$\Phi = -EA = -\frac{qR^2}{4 \epsilon_o z^2}.$$
I think this is correct, but is this not assuming the disk is of negligible thickness? If the flux penetrates the disk and then leaves, the net flux would then be zero.

The charge on the final sphere (drop) is 27x as much as a single drop. The volume of one drop is 4/3 π R3 and that of the final drop is 4/3 π R'3. I think the volume of the final drop would be greater than that of a single drop, but I am still unsure of how you find the radius. I was thinking that as you add more drops, the radius will increase by 1/R each time => when you add 27 drops, the radius is R + 27/R. It makes sense that the radius will not actually increase by 27 x as much since we are not sticking them together but rather coalescing them.

Last edited: Feb 3, 2013
7. Feb 3, 2013

### CAF123

I think the question wants a non zero approx, like you said, so should I make the assumption that we can neglect the thickness of the disk?

8. Feb 3, 2013

### TSny

Yes, I think so.

9. Feb 3, 2013

### CAF123

Thanks TSny. Are all my results correct now for both questions?

10. Feb 3, 2013

### TSny

For the second question you still need to get the correct radius of the large sphere. The density of water does not change when you combine the drops. Also, since mass must be conserved, how does the mass of the large sphere compare to the mass of one of the drops.

Then think about the relation: mass = density x volume. That should help to determine how the volume of the large sphere compares to the volume of one of the drops. Since you know how volume depends on radius, you can then determine how the final radius compares to the radius of a drop.

11. Feb 3, 2013

### TSny

12. Feb 4, 2013

### CAF123

I got that the new sphere has a radius 3x that of the original with a new potential $$V_{new} = \frac{27q}{4 π ε_o 3R} = \frac{9q}{4 π ε_o R}$$ so charge gets distributed 9 times as much as the radius increases.

Is there a reason why simply adding all the potentials together is wrong. I.e just doing $$\frac{q}{4 π ε_o R} +\frac{q}{4 π ε_o R} + \frac{q}{4 π ε_o R}?..,$$ etc 27 times?

Last edited: Feb 4, 2013
13. Feb 4, 2013

### TSny

That looks good.
Not sure how to interpret that statement. But the result of the potential of the new sphere is the same as the potential of one of the little drops if it had a charge of $9q$.
If you added 27 of those terms, you would get $$\frac{27q}{4 π ε_o R}$$ which has the correct amount of charge, but it doesn't take into account the increase in radius.

14. Feb 4, 2013

### CAF123

Is there a reason why that method is wrong? Would it be correct to say that simply adding 27 of those terms would give the potential of a single drop if 27 charges were all at an equidistant distance from this initial drop. My original method therefore neglects that the drops are actually coelescing.

15. Feb 4, 2013

### TSny

Adding the 27 terms is equivalent to taking all the charge of the 27 drops and putting all that charge on just one of the little drops of radius R. So, as you say, it doesn't take into account the increase in radius of the sphere as the drops coalesce.

16. Feb 4, 2013

### CAF123

Many thanks TSny