# A fly between capacitor plates

1. Apr 8, 2013

### Saitama

1. The problem statement, all variables and given/known data
(see attachment)

2. Relevant equations

3. The attempt at a solution
I don't understand the question. There are equipotential lines everywhere in the region between the capacitor plates, even at $r_0$ from the left plate. So the fly could simply fly along that line and the answer should be $\frac{d\epsilon}{2}$. Can anyone tell me if I am interpreting the question correctly?

Any help is appreciated. Thanks!

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2. Apr 8, 2013

### voko

You are supposed to find what the equipotential surface looks like, and its maximum distance from the capacitors center. Note they are talking about lines, not surfaces. Which is the same thing because of the symmetry in the problem.

3. Apr 8, 2013

### Saitama

I still have no idea about this problem. Won't the equipotential lines look like as they are shown in the following figure:

4. Apr 8, 2013

### TSny

You need to consider what the equipotential lines look like when extended out from the region between the plates in order to get an idea of the path the fly will travel.

5. Apr 8, 2013

### Saitama

I don't know if I understand it right but do you mean that I need to consider the lines as 3-D planes and assume that fly travels on a parabolic path on that 3-D plane situated at the distance $r_0$?

6. Apr 8, 2013

### TSny

See the equipotential lines in the attachment. The surfaces would be obtained by rotating the figure about an axis perpendicular to the plates and passing through the center of the capacitor.

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7. Apr 8, 2013

### Saitama

So I should not consider the given capacitor as a parallel plate capacitor where the electric field is constant?

I still have no idea about how should I begin making equations.

8. Apr 8, 2013

### voko

Just compute the potential explicitly. Use the symmetry, it should be some nice function.

9. Apr 8, 2013

### Saitama

At what point should I calculate the potential? On the axis perpendicular to the capacitor plates and passing through their centre?

10. Apr 8, 2013

### voko

Everywhere (but the symmetry should simplify that). Make it a function of the vector distance from the center of the capacitor.

11. Apr 9, 2013

### Saitama

I don't see the symmetry you are talking about. Could you please make a diagram or a sketch? That would help a lot.

12. Apr 9, 2013

### voko

The plates are discs. So the symmetry is radial. You only need to consider the potential in a plane passing through the centers of the disks.

13. Apr 9, 2013

### haruspex

It's rather easier than that.
Pranav-Arora, from TSny's diagram, do you understand the approximate shape of the equipotential surface? Note that it is topologically a sphere, and has an axis of symmetry the same as that of the capacitor. Where do you think it gets furthest from the centre of the capacitor? You only need to compute the potential for such points.

14. Apr 9, 2013

### voko

That assumes either taking the diagram for granted, or being able to explain why it should look so.

15. Apr 9, 2013

### haruspex

Agreed, and I anticipated that would arise in discussion. But I believe it is relatively easy to explain it qualitatively rather than explicitly evaluate the potential everywhere.

16. Apr 10, 2013

### Saitama

Are you talking about the field lines just near the centre of capacitor?

I honestly tried but I couldn't set up the integrals. I tried to find the potential at point P but I have no clue how would I set up the integrals.

#### Attached Files:

• ###### disk.png
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17. Apr 10, 2013

### ehild

Look at TSny picture showing equipotential lines. I copied it here and added the one that goes exactly halfway between the plates. If the fly flew along that plane, it would reach infinity. All other equipotential surfaces are closed and reach farthest along the axis of the capacitor. (You can refer to symmetry). You find the electric field of a charged plate along its axis rather easily, http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html, now you have two disks with opposite charge, and close together. Find the distance where the potential is the same as at the starting position of the fly.

ehild

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• ###### Fly.JPG
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18. Apr 10, 2013

### haruspex

No, equipotentials are orthogonal to field lines. Field lines follow the direction of greatest rate of change of potential, whereas equipotentials follow the directions of no change in potential. In 3D, equipotentials are surfaces, often topologically spherical. See TSny's diagram.
It should be reasonably obvious whereabouts these surfaces get furthest from the centre of the capacitor. (This will be well outside of it.) As voko points out, this observation should be supported by some reasoning, but I think that can be done without having to perform any integrals.
Once you've determined the line on which this point must lie, it's a simple matter to determine whereabouts on the line the point will be.

19. Apr 11, 2013

### Saitama

But how would I find the potential at the initial position of the fly?

20. Apr 11, 2013

### ehild

Assume the fly starts from the axis of the capacitor. Assume the charge of the capacitor is Q. And the distance between the fly and the positive plate is r0=d/2(1-ε). d=1 cm, ε=10-4 (problem text).

ehild

Last edited: Apr 11, 2013