# Electrostatic question with three circular disks.

1. Jun 20, 2011

### dHannibal

1. The problem statement, all variables and given/known data
Two discs are each charged +q. A third metal disc of radius R∗ > 5
cm is carefully inserted between the two discs; the third disc is neutral and is suspended by
an electrically insulating thread. The three discs are all parallel to each other and their
centers lie along the same horizontal line ,so that when viewed head-on the discs are
concentric circles. The resulting set-up is shown in the figure.

Find the radius R∗ of the third disc such that the net electrostatic force acting on each
charged disc is zero. (The fringing effect is neglected in this problem).

[PLAIN]http://img89.imageshack.us/img89/4135/vdcx.png [Broken]

2. Relevant equations

The charge distribution on a single charged disc is
$$\sigma _{\text{disk}} = \frac{Q}{4 \pi R \sqrt{R^2 - r^2}}.$$

3. The attempt at a solution
For starters,I can't figure out how the charges will be distributed on the neutral disc.

Last edited by a moderator: May 5, 2017
2. Jun 20, 2011

### bjnartowt

Because the problem statement says to neglect fringing, it may be possible to treat the disks as "infinite planes". The electric field of an infinite plane is well-known. I'm further convinced that the "infinite plane" solution may be used because of the symmetry with which the disks are arranged (such that if they were transparent, somehow, they'd be seen as concentric circles).

I think the charge density will be found by a simple area-fraction. Metal objects always have their charges move about such that fields are perfectly-cancelled out within their volume. Treat that as your charge distribution, sigma.

You can then compute electrostatic pressure with this known sigma, which, in turn, could be integrated over an area to constitute forces that you could balance to zero.

Does that help at all? This doesn't look like a "Griffiths electrodynamics" problem I've seen, but that's the textbook I'm familiar with.

3. Jun 22, 2011

### Philip Wood

Both faces of C will have charge -q induced on them in the central region where they're sandwiched between A and B. So their outer zones (where not between A and B) will have charges +q on each face (since C is, overall, neutral, or, putting it another way, free electrons will be pulled, by forces from A and B, into C's inner region, leaving C's outer zones depleted of free electrons).

A, being positive, will experience a force, F, towards C, due to the negative charge on C's central region, but also a force, G, away from C, due to the positive charges on C's outer zones. [What goes for A also goes for B]

As I see it, G will never be as large as F, because the charge on the outer zones of C is further from A and is pulling at an angle. [Incidentally, I believe that we need to count the total charge, +2q, on both faces of the outer zones, but I wouldn't think this would be enough to make G as large as F.] So I don't see that the force on A could be zero, whatever the radius of C. I'm not at all sure that we can use the formula you quote, because this doesn't seem to take account of the different environments of the sandwiched and 'free' parts of the disc. Unless, perhaps, the formula is supposed to apply just to the outer zones of C. Is it?]

Last edited: Jun 22, 2011
4. Jun 22, 2011

### bjnartowt

I have some questions,

1) What kind of textbook did this question come up in--a graduate electrodynamics text, or undergraduate?

2) Are the radii of disks A and B given? (They appear not to be, as I'm sure you copied the problem statement verbatim, but I thought I'd ask anyway). If they are given, the idea I originally had might work (in which the charge induced on the faces of C are *not* trivially + q. If not, it's back to the drawing board for me...

5. Jun 24, 2011

### dHannibal

This was in a physics problem book with various levels of problems,so I am not sure which level it is. But probably undergrad.
I also tried to apply phil's model,and it seems that equilibrium is not possible in this approach. The formula is the charge distribution on a free circular disc,and as the fringing effects are neglected,I believe it could be applied to the outer regions on the large disc.
bjnartowt,yes the radii are given as 5 cm.

Last edited: Jun 24, 2011
6. Jun 24, 2011

### bjnartowt

No, I meant the two larger disks. The smaller disk is 5 cm radius, but I guess it appears the larger disk doesn't have its dimensions specified?

7. Jun 24, 2011

### dHannibal

bjnartowt,there is only one large disc,and two smaller discs. The question asks the radius of the large disc for the equilibrium state,if there is one.

8. Jun 26, 2011

### fizika_kz

heyyy. I remember this problem. This problem is from World Physics Olympiad, isn't it dHannibal?