Electrostatic question with three circular disks.

In summary, 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 net electrostatic force acting on each charged disc is zero.
  • #1
dHannibal
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Homework Statement


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

Homework Equations



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

The Attempt at a Solution


For starters,I can't figure out how the charges will be distributed on the neutral disc.
 
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  • #2
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
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?]
 
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  • #4
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
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.
 
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  • #6
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
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
heyyy. I remember this problem. This problem is from World Physics Olympiad, isn't it dHannibal?
 

FAQ: Electrostatic question with three circular disks.

1. How do the charges on the three circular disks interact with each other?

The charges on the three disks interact through the electrostatic force, which follows the principle of like charges repelling each other and opposite charges attracting each other.

2. What determines the strength of the electrostatic force between the disks?

The strength of the electrostatic force between the disks is determined by the magnitude of the charges on each disk and the distance between them, according to Coulomb's Law.

3. Can the charges on the disks be both positive or negative?

Yes, the charges on the disks can be both positive or negative, as long as they are not all the same. For example, if one disk has a positive charge and the other two have negative charges, or vice versa.

4. How does the distance between the disks affect the electrostatic force?

The electrostatic force decreases as the distance between the disks increases. This is because the force follows an inverse square law, meaning that as the distance doubles, the force decreases by a factor of four.

5. Is there a limit to the number of disks that can be included in an electrostatic question?

No, there is no limit to the number of disks that can be included in an electrostatic question. However, as the number of disks increases, the calculations become more complex and the results may not be as straightforward.

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