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Say you have a hollow sphere with a uniformly distributed charge on the surface. Why is the electric field everywhere inside the sphere zero?

For the centre, its easy to add the vectors from the surface charges and show they sum to zero because of symmetry.

But can you show me how the field cancels out to zero for points other than the centre using vector addition?

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  • $\begingroup$ It occurs to me that your question is a little ambiguous as stated. Are you asking about the field inside a good conductor (zero even in the presence of an external field) or the field due to a spherical symmetric distribution (zero at all radii lower that the smallest radius with charge). $\endgroup$ Commented Jan 24, 2014 at 3:08
  • $\begingroup$ @dmckee I'm asking about the field due to a spherical symmetric distribution. $\endgroup$
    – dfg
    Commented Jan 24, 2014 at 3:09

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It is very simple, but strictly applies to a conducting sphere.

From ANY point inside a sphere, draw a double cone that is zero dimension at that point.

The cross-section of the cone, can be any shape.

The two ends of the cone intersect the sphere in two similar shaped curved surfaces. Any line from a point inside one of those areas, through the tip of the cone to the opposite surface has two sections with a ratio of A:B in length.

The areas of the two end cap surfaces, are also in the ratio of A^2:B^2 and so are the charges on those two areas.

Since the inverse distances squared are also A^2:B^2, the force on a charge at the cone tip is net zero. This is true for any point in the sphere, and any cone angle or cross-section shape.

QED

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    $\begingroup$ It applies perfectly well to any spherically symmetric distribution of charge. Conductors or insulators have nothing fundamental to do with it, though using a conductor allows you to easily guarantee the spherical symmetry. $\endgroup$ Commented Jan 24, 2014 at 2:53
  • $\begingroup$ galileo.phys.virginia.edu/classes/152.mf1i.spring02/… $\endgroup$
    – dfg
    Commented Jan 24, 2014 at 2:57
  • $\begingroup$ The link above uses the same argument (I think) to show that the gravitational field inside a sphere is always pointing towards the centre. $\endgroup$
    – dfg
    Commented Jan 24, 2014 at 2:58
  • $\begingroup$ Also I don't really understand what you did after recognizing the areas of both ends are the same. Could you maybe add in diagrams to make the proof more obvious? $\endgroup$
    – dfg
    Commented Jan 24, 2014 at 2:59
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The record that Virginia.edu explanation of the direction of gravity, is simply awful, although it gets the correct answer. The direction is obvious, once it is understood that the gravity inside a shell is zero, so only the smaller sphere inside produces a gravity field, which can only point to the cg of the sphere.

Any surface that is perfectly NON-CONDUCTING can not have electric charges moving on its surface.

Therefore it is impossible to put a uniform distribution of charge, on a non-conducting sphere.

My proof is ONLY valid for a conducting surface.

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  • $\begingroup$ It appears that you would really be better off using a registered account and reading the help center page on merging accounts. You have at least 6 accounts using this name (various capitalization) and gravitar. Without using a consistent account you will never achieve the privilege level needed to make full use of the site, and we will delete any comments posted as answers. $\endgroup$ Commented Jan 24, 2014 at 19:29
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Because any Gauss surface made inside the hollow sphere encloses no charge so the net flux is zero. And if the Gaussian surface is shrunk to zero dimension, the electric field is zero at that point.

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    $\begingroup$ I specifically requested an explanation using vector addition! $\endgroup$
    – dfg
    Commented Jan 24, 2014 at 2:07
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    $\begingroup$ In any case, this explanation is incomplete without an argument from symmetry to bridge the gap from zero flux to zero field. $\endgroup$ Commented Jan 24, 2014 at 2:54
  • $\begingroup$ @dmckee How does zero flux in a Gaussian surface imply the field inside is zero (assuming symmetry)? $\endgroup$
    – dfg
    Commented Jan 24, 2014 at 3:01
  • $\begingroup$ @dmckee Because if you have an empty sphere and a field passes through it, the flux would be zero (since the field "flows" out the other side) but the field inside wouldn't be zero. $\endgroup$
    – dfg
    Commented Jan 24, 2014 at 3:02

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