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If you have a conducting hollow sphere with a uniform charge on its surface, then will the electric field at every point inside the shell be 0.

The reason the electric field is 0 at the center is clear from the symmetry of the sphere, but for a point at a certain distance from the center, shouldn't a net electric field exist?

I saw a similar question on the site but it was vaguely explained.

Qmechanic
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Tabish Mir
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  • Possible duplicate: https://physics.stackexchange.com/q/150238/2451 Related: https://physics.stackexchange.com/q/571402/2451 and links therein. – Qmechanic Nov 23 '21 at 04:10

2 Answers2

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If the shell and its charge distribution are spherically symmetric and static (which your question does imply when you say "uniform charge"), and if electric field lines begin and end on charges, then we know that any electric field that might be present inside the shell must be directed radially (in or out, i.e. $E_{\theta} = E_{\phi}=0$).

From there, a simple application of Gauss's law, using a spherical surface centered on the center of the shell tells you that the radial electric field component must also be zero at any radial coordinate $r$ within the sphere. $$ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0} = 0$$ $$ 4\pi r^2 E_r = 0$$ $$\rightarrow E_r = 0 $$

Therefore, we can say that at any point within the sphere (defined by $r$ and two angular coordinates) that $E_r = E_{\theta} = E_{\phi}=0$ and so the total electric field at any point (inside the sphere) is zero, not just the centre.

ProfRob
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    This is not a correct answer, since it doesn't explain at all, why an off-center point inside the sphere has zero charge. The fact that the sphere is "spherically symetric and static" is quite obvious but the OP was asking specifically how does this imply the zero charge. Gauss's law only describes electric field originating in an enclosed region – and again, it's obvious that an empty patch of space wouldn't generate any electric field. – m93a May 30 '18 at 07:53
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    @m93a It is the correct answer, since it demonstrates that the electric field is zero everywhere inside the shell - which is what is asked. The question does not ask about the charge distribution, which is specified to be "on its surface". – ProfRob May 30 '18 at 08:05
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    I find this answer unsatisfactory. The OP mentions the surface charge and makes pains to remind us that the distribution is uniform, and he mentions the obvious cancellation at the center. – garyp May 30 '18 at 19:11
  • @garyp Huh? I'm finding the criticism of my textbook answer rather difficult to fathom. What do you think the question is about? I think it is asking for a demonstration/argument that the electric field is everywhere zero inside a shell with a symmetric charge distribution, and not just at the centre. That is what I have provided. – ProfRob May 30 '18 at 20:16
  • Symmetry is not essential for the electric field to vanish throughout the interior. The shell merely needs to be closed. The inner surface of the conducting shell must be an equipotential, lest the charge on the shell redistribute. Given that the shell is closed, this guarantees that the potential will be constant everywhere inside. – Bert Barrois May 30 '18 at 20:46
  • I don't understand the criticism either because this clearly answers the OP's question which is very specific about the charge distribution and geometry of the conductor. The proof is more difficult for an arbitrary hollow conductor of any geometry with excessive charge on its surface, but this isn't what was asked. – Physiks lover May 30 '18 at 21:04
  • I find the criticism and downvotes of this answer mystifying (being polite). It is telling that there is one alternative answer proposed, that says exactly the same thing (but doesn't prove it) with no votes, up or down – ProfRob May 03 '19 at 09:02
  • How do you justify Er=Eθ=Eϕ=0 ? I understand the system from a uniqueness theorem point of view, as well as the idea of hollowing out a charged solid sphere to make a hollow sphere and the surface charge doesn't change, thus not changing E=0. But I don't see an easy way to justify non-radial components of E = 0 just from a plain Gauss's Law argument. – lamplamp Jan 10 '21 at 09:05
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    @lampkamp Do you mean how can I justify that the non-radial components of the E-field are zero? This is covered by the first sentence of my answer. How can you generate a non-radial E-field from a spherically symmetric charge distribution about the origin? Gauss's law only tells you about the radial component in this case. – ProfRob Jan 10 '21 at 10:07
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if you consider an off-center point,the field created by the charges near to the point should be equal to the field created by other charges sitting opposite to it. Amount of charge and distance varies will prove this field or field strength is same. As they are opposite the net field is zero at any off-center point.

STA
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