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We know from electrostatics that the field of an infinite sheet of surface charge density $\sigma$ is

$$ \vec E = \frac{\sigma}{2\epsilon}\hat{x} $$

But the field inside the capacitor is

$$ \vec E = \frac{\sigma}{\epsilon}\hat{x} $$

That, in my understanding, is due to the addition of positive and negative plates fields

However in any other geometry such as a cylindrical geometry, we use the field due to one plate only. For instance, the following:

Capacitance inside a Cylinder

The E field is identical to that of only one plate, not the addition of the positive and negative plates fields. Why is this the case?

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    $\begingroup$ Hi and welcome to the Physics SE! The equations become much easier to read, search and edit when mathjax is used. I've proposed an edit to your post this time, but you should use it yourself in your next posts. $\endgroup$
    – Styg
    Dec 23, 2017 at 10:38
  • $\begingroup$ I searched for it in the editor, but without avail. $\endgroup$
    – Raafat Abualazm
    Dec 23, 2017 at 10:39
  • $\begingroup$ Yeah, I see the editor only has Markdown help. You might find this help page useful too. $\endgroup$
    – Styg
    Dec 23, 2017 at 10:43
  • $\begingroup$ More on capacitors and factors of 2: physics.stackexchange.com/q/110480/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Dec 23, 2017 at 12:19

1 Answer 1

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We do take into account the field due to both plates in the cylindrical geometry. The field due to an infinite uniformly charged cylinder is zero in the interior of the cylinder, as can be shown by symmetry arguments combined with Gauss's Law.

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  • $\begingroup$ How come it is zero? When drawing Gaussian surface, there's clearly a field in the radial direction. $\endgroup$
    – Raafat Abualazm
    Dec 23, 2017 at 10:33
  • $\begingroup$ Right, the field, if any, will always be in the radial direction due to cylindrical symmetry, and we always take a cylinder with the same axis as a Gaussian surface. But a Gaussian surface drawn inside a charged cylinder will not enclose any charge, which means that the field must be zero at the Gaussian surface. $\endgroup$
    – Styg
    Dec 23, 2017 at 10:40
  • $\begingroup$ But the Gaussian surface is between both conducting cylinders, so there is field in between. My question is why in such a geometry the field inside is equivalent to the field of any of the plates, however in the parallel plate it is double that of any individual plate. All geometries share this feature except the parallel plate. $\endgroup$
    – Raafat Abualazm
    Dec 23, 2017 at 11:00
  • $\begingroup$ "All geometries share this feature except the parallel plate" Not true. Imagine two uniformly charged balls with equal and opposite charge placed a distance apart in vacuum. Nowhere is the field that of a single charged ball. $\endgroup$
    – Styg
    Dec 23, 2017 at 11:41
  • $\begingroup$ The Electric field, as calculated by Gauss' law, is identical whether there is a negatively charged plate or not. $\endgroup$
    – Raafat Abualazm
    Dec 25, 2017 at 11:48

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