How Does a Fly Navigate Between Capacitor Plates?

[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+<br /> d^2/4}\sqrt{1-\frac{xd}{x^2+R^2+<br /> d^2/4}}...

ehild

 

[Saitama]

\sqrt{x^2+\frac{d^2}{4}-xd+R^2}=\sqrt{x^2+R^2+<br /> d^2/4}\sqrt{1-\frac{xd}{x^2+R^2+<br /> 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!

 

[ehild]

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

 

[Saitama]

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]

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