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If I take a spherical shell, then give it a uniform charge somehow, the E-field lines will only point outwards, right?

Then what about for a square thin surface that is initially laid flat, charged, and then bent into a cylinder (with open ends)? Which way do the E-field lines point then? Before it was bent, the surface had E-field lines pointing up and down. If you bend it into a cylinder, I would suspect that the E-field lines continue to point inwards.

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The case is that the electric field is always normal to the surface of a conductor. This is a consequence of the fact that charges on the surface will continue to move until there is no net force on any of the charges. At that moment, the only direction the field can point in is the "outward" normal direction.

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  • $\begingroup$ Wait, so then you're saying that my first part is wrong? The E-field lines of a sphere point both inward and outward for an empty spherical shell (infinitesimally thin) of uniform charge? $\endgroup$ Sep 10, 2016 at 17:58
  • $\begingroup$ @whatwhatwhat inside any conductor, there is zero electric field. That includes inside of cavities like a hollow ball. $\endgroup$
    – Alex Ortiz
    Sep 10, 2016 at 18:00
  • $\begingroup$ But suppose that I take this spherical shell and open it up so as to lay it flat. Do the E-field lines reorient themselves so as be normal to the top and bottom? Furthermore, what if I then re-assemble it into a sphere again? Is there still zero E-field inside? $\endgroup$ Sep 10, 2016 at 18:02
  • $\begingroup$ @whatwhatwhat yes, if you were to open it up, the electric field would resume emerging from each side of the now not-closed surface $\endgroup$
    – Alex Ortiz
    Sep 10, 2016 at 18:36

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