0
$\begingroup$

I'm grappling with understanding Gauss' Law as applied to charged sheets and oppositely charged plates.

From what I've gathered, when using a Gaussian pillbox encompassing both sides of an infinite sheet, we can derive the electric field to be $2EA=\frac{\sigma A}{\epsilon_0}$​, which simplifies to $E=\frac{\sigma}{2\epsilon_0}$.

Now, if I consider two oppositely charged plates, I understand that the resultant field outside both plates would be zero. However, I'm uncertain about the field within the plates.

Could it be conceptualized by taking a Gaussian surface only encompassing half the height of each plate and, consequently, accounting for only half the charge on each plate? Would this approach result in $AE=\frac{(\sigma/2)A}{\epsilon_0}$​, and thus $E=\frac{\sigma}{2\epsilon_0}$​ for each plate? Consequently, would the electric field inside both plates then become $E=\frac{\sigma}{\epsilon_0}$?

And what about if my plates are perfectly conducting? Then the charge at the top of a plate and bottom could be different right? This would also affect the resulting outcome if a charge is induced on them?

I'm seeking clarification on this interpretation of Gauss' Law in this scenario. Any insights or corrections would be greatly appreciated.

$\endgroup$

1 Answer 1

0
$\begingroup$

The case of infinite sheets of charge is a problem amenable to Gauss' law, however, a set of finite sized plates is not. The finite size of the plates destroys the symmetry needed to apply Gauss' law, so finding the field in between and outside the plates is not so simple a project.

Is there a field outside the plates? Of course there is. In fact, it is a field such that the further and further you get from the plates, the more and more the field looks like that of a point dipole. Just because the flux through a spherical closed surface surrounding the plates is zero, as it must be since the net enclosed charge is zero, doesn't mean that the field inside is zero. It just means that there are as many lines of flux leaving the sphere as are incoming to the sphere.

As far as the field in the region between the plates, if you are far enough away from the edges of the plates, so that these can be ignored, then the field is given by the expression for two oppositely charged infinite sheets. This is the approximation often used to model the field for plate capacitors. In reality, however, the field is more complicated. To get an idea of how to handle the effects of the edges, you may want to consult a graduate level text like J.D. Jackson's Classical Electrodynamics.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.