How Does a Fly Navigate Between Capacitor Plates?

[Saitama]

Homework Statement

(see attachment)

Homework Equations

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!

 

[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.

 

[Saitama]

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.

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

 

[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.

 

[Saitama]

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.

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 $r_0$?

 

[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.

 

[Saitama]

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.

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.

 

[voko]

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

 

[Saitama]

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

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

 

[voko]

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

 

[Saitama]

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

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

 

[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.

 

[haruspex]

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

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.

 

[voko]

It's rather easier than that.
Pranav-Arora, from TSny's diagram, do you understand the approximate shape of the equipotential surface?

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

 

[haruspex]

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

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.

 

[Saitama]

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.

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

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.

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.

 

[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

 

[haruspex]

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

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.

 

[Saitama]

Find the distance where the potential is the same as at the starting position of the fly.

ehild

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

 

[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

 

[Saitama]

Can you find the electric field between capacitor plates?

ehild

Yes, I can calculate the electric field at any point on the axis. The expression for the electric field I get is a bit messy. The direction is towards the negative plate. Should I use the approximation that the distance between the plates is much smaller than the radius of capacitor plates? If so, I get the expression:
$$E=2\pi k \sigma \left(2-\frac{d}{R}\right)$$

 

[ehild]

You need the potential really. What is it at distance r0 from the positive plate?

ehild

 

[Saitama]

You need the potential really. What is it at distance r0 from the positive plate?

ehild

The formula for electric potential of a charge disk at any point on the symmetrical axis is:
$$V=2\pi k \sigma(\sqrt{x^2+R^2}-x)$$
where x is the distance from the centre of disk.
At a distance $r_0$ from the positive plate is,
$$V=2\pi k \sigma(\sqrt{r_0^2+R^2}-r_0)-2\pi k \sigma(\sqrt{(d-r_0)^2+R^2}-(d-r_0))$$
$$V=2\pi k \sigma(\sqrt{r_0^2+R^2}-\sqrt{(d-r_0)^2+R^2}-2r_0)$$
Should I use the approximation d<<R?

 

[ehild]

Express ro with d and ε and use that ε<<1.


ehild.

 

[Saitama]

Express ro with d and ε and use that ε<<1.

$$\sqrt{r_0^2+R^2}=\sqrt{\frac{d^2}{4}(1+\epsilon^2-2\epsilon)+R^2}$$
Since $\epsilon <<1$
$$\sqrt{r_0^2+R^2}=\sqrt{\frac{d^2}{4}+R^2-\frac{d^2\epsilon}{2}}$$

Similarly,
$$\sqrt{(d-r_0)^2+R^2}=\sqrt{\frac{d^2}{4}+R^2+\frac{d^2\epsilon}{2}}$$

I don't know what should I do next?

 

[ehild]

You can factor out R²+d²/4 from both square roots, and then use the approximate formula sqrt(1+β)=1+β/2 for β<<1

ehild

 

[Saitama]

You can factor out R²+d²/4 from both square roots, and then use the approximate formula sqrt(1+β)=1+β/2 for β<<1

ehild

Using the approximation,
$$\sqrt{r_0^2+R^2}-\sqrt{(d-r_0)^2+R^2}=-\frac{d^2\epsilon}{2}\left(R^2+\frac{d^2}{4}\right)^{-\frac{1}{2}}$$

What am I supposed to do now?

 

[haruspex]

$$V=2\pi k \sigma(\sqrt{r_0^2+R^2}-r_0)-2\pi k \sigma(\sqrt{(d-r_0)^2+R^2}-(d-r_0))$$
$$V=2\pi k \sigma(\sqrt{r_0^2+R^2}-\sqrt{(d-r_0)^2+R^2}-2r_0)$$

You seem to have dropped a term.. Shouldn't it be:
$V=2\pi k \sigma(\sqrt{r_0^2+R^2}-\sqrt{(d-r_0)^2+R^2}+d-2r_0)$?

$$\sqrt{r_0^2+R^2}-\sqrt{(d-r_0)^2+R^2}=-\frac{d^2\epsilon}{2}\left(R^2+\frac{d^2}{4}\right)^{-\frac{1}{2}}$$

$V=2\pi k \sigma(-\frac{d^2\epsilon}{2}\left(R^2+\frac{d^2}{4}\right)^{-\frac{1}{2}}+d-2r_0)=2\pi k \sigma(1-\left(1+\frac{4R^2}{d^2}\right)^{-\frac{1}{2}})d\epsilon$
So far so good, but now you need the potential a long way from the centre, distance x say.
$V_{ext}=2\pi k \sigma((\sqrt{(x-d/2)^2+R^2}-x+d/2) - (\sqrt{(x+d/2)^2+R^2}-x-d/2))$
Approximate that for large x.

 

[ehild]

Well, that is the method of simplifying the roots you can apply through the derivation. If you set up the potential correctly, you will reach to the correct solution. But for that, you need to fix the point of reference and write up the potential with respect to it, in terms of a coordinate. The problem asked how far the fly reaches from the middle point between the capacitor plates. Put the origin there and choose the x-axis along the line connecting the centres of the plates.

Be careful with the expression of the potential. x was the distance from the plate but the electric field has opposite signs at both sides of the plate. The correct formula is $$V=2 \pi k \sigma(\sqrt{x^2+R^2}-|x|)$$ You need to express that "x" for both plates in terms of the common coordinate. Your potential formula gave a negative potential value for a point which was closer to the positively charged plate than to the negatively charged one -with respect to what was it negative?



ehild

 

[ehild]

I suggest a very rough approximation but it seems to work. (At last, I got the same result with it as with the "exact" method).

The plates are close enough with respect to the diameter, so you can treat the set-up as an ideal capacitor when determining the electric field and the potential between them, with respect to the middle point.

Very far away from the capacitor the plates are small with respect to the distance so they can be treated as point charges. Determine how the potential depends on the distance from the central point.

See picture. The fly starts at P and reaches P' on the equipotential.
By the way, you had the electric field between the plates correctly in post #21. What is the potential at the point εd/2 distance from the centre, if you take the zero of the potential at the central point? You can even omit the d/R term.ehild

 

[Saitama]

Very far away from the capacitor the plates are small with respect to the distance so they can be treated as point charges. Determine how the potential depends on the distance from the central point.

The expression for the potential outside is mentioned above by haruspex.
For large x, the expression approximates to:
$$V=2\pi k \sigma d$$
$$\Rightarrow V=\frac{\sigma d}{2 \epsilon_0}$$

Have I approximated the expression correctly?

See picture. The fly starts at P and reaches P' on the equipotential.
By the way, you had the electric field between the plates correctly in post #21. What is the potential at the point εd/2 distance from the centre, if you take the zero of the potential at the central point? You can even omit the d/R term.

So the electric field in the middle region can be approximated to $\displaystyle \frac{\sigma}{\epsilon_0}$?

The potential at the point εd/2 distance from the centre is $\displaystyle \frac{\sigma}{\epsilon_0}\cdot \frac{d\epsilon}{2}$

 

[ehild]

The expression for the potential outside is mentioned above by haruspex.
For large x, the expression approximates to:
$$V=2\pi k \sigma d$$
$$\Rightarrow V=\frac{\sigma d}{2 \epsilon_0}$$

Have I approximated the expression correctly?

No, it must depend (decrease) with the distance from the capacitor.

So the electric field in the middle region can be approximated to $\displaystyle \frac{\sigma}{\epsilon_0}$?

The potential at the point εd/2 distance from the centre is $\displaystyle \frac{\sigma}{\epsilon_0}\cdot \frac{d\epsilon}{2}$

That is correct.


ehild

 

[Saitama]

No, it must depend (decrease) with the distance from the capacitor.

I am not sure how would I approximate the following surd.
$$\sqrt{\left(x-\frac{d}{2}\right)^2+R^2}=\sqrt{x^2+\frac{d^2}{4}-xd+R^2}$$

I guess I can drop the d^2/4 term but what should I do after that?

 

[ehild]

$$\sqrt{x^2+\frac{d^2}{4}-xd+R^2}=\sqrt{x^2+R^2+ d^2/4}\sqrt{1-\frac{xd}{x^2+R^2+ d^2/4}}$$...

ehild

 

[Saitama]

$$\sqrt{x^2+\frac{d^2}{4}-xd+R^2}=\sqrt{x^2+R^2+ d^2/4}\sqrt{1-\frac{xd}{x^2+R^2+ d^2/4}}$$...

ehild

$$\sqrt{x^2+R^2+d^2/4}\sqrt{1-\frac{xd}{x^2+R^2+d^2/4}}=\sqrt{x^2+R^2+d^2/4}\left(1-\frac{xd}{2(x^2+R^2+d^2/4)}\right)$$

Similarly, the other surd can be simplified.
I end up with the following expression for the potential:
$$V=2\pi k \sigma d \left(1-\frac{x}{\sqrt{x^2+R^2+d^2/4}}\right)$$

Is this correct?

EDIT: Looks like I have got the right answer. Equating the above expression for $\frac{\sigma d \epsilon}{2 \epsilon_0}$ and plugging in the values of d and R, I get the right answer which is 21.215.

Thanks a lot for the help everyone!