0
$\begingroup$

Say there is a conducting cylinder in a uniform external field. The charges on the cylinder surface will distribute and create a surface charge density.

Is there any reason the charge density would generally be uniform across the surface? We know: $$\sigma = \Delta E \cdot \varepsilon_0 \cdot \hat{n}$$ But what physically ensures the electric field just outside the conductor is equal in magnitude all around the encompassing circle?

$\endgroup$
  • 1
    $\begingroup$ Charge density can't be uniform across the surface in this case. If originally the total charge of conductor was 0 it will remain 0 after we put the conductor into external field. Charges can't accumulate inside the conductor, so to total charge of the surface is still zero. And if density is uniform - it must be zero everywhere. $\endgroup$ – lesnik May 27 at 6:49
0
$\begingroup$

I misread the question and so have rewritten my answer.

The surface will be an equipotential but to maintain it as an equipotential in an external field charges will have to move over the surface and they are called induced charged.

The diagram below illustrates this.

enter image description here

You will note by the uneven spacing of the field lines that the electric field around the conductor is not uniform.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Does that mean the magnitude of the electric field "just outside" will be the same all around? How does that go hand in hand with lesnik's comment? $\endgroup$ – Darkenin May 27 at 6:58
  • $\begingroup$ @Darkenin I misread the question and so have rewritten my answer. $\endgroup$ – Farcher May 27 at 10:20
  • $\begingroup$ The electric potential must be the same at every point inside and just outside of the conductor. That means that the E field cannot have a component parallel to the surface outside at the surface. Referring to the sketch, E would have to be close to zero near the top and bottom of the conductor. $\endgroup$ – R.W. Bird May 27 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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