# A fly between capacitor plates

ehild
Homework Helper
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

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

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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?

ehild said:
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
Homework Helper
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

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

ehild

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

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! Last edited:
ehild
Homework Helper
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! Very good!

Try also the method I outlined in #30. Just for fun :)

ehild

Try also the method I outlined in #30. Just for fun :)
That's the method I used to solve the problem. Are you talking about #29?

ehild
Homework Helper
That's the method I used to solve the problem. Are you talking about #29?
No it was #30. Approximating the field far away as that of two points charges.

ehild