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For a conducting sphere of radius $a$ and charge $Q$ surrounded by a concentric spherical shell of radius $b$ and charge $-Q$ (where $b > a$), I am aware that the electric field in the volume between the surface of the included sphere to the outer spherical shell would be:

$$E = \frac{Q}{4\pi\epsilon_0r^2}$$

where $r$ is the distance away from the center (of both the sphere and the shell) by Gauss's Law.

The question is isn't this exactly the same as the electric field of an isolated sphere, where the outer shell doesn't exist? Does the negatively charged outer shell not add to the electric field? Why is this the case?

Please ask me any questions regarding the problem (I'm in 9th grade and my explanations might be confusing)

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  • $\begingroup$ Are you familiar with the argument of why the electric field inside a charged hollow sphere is zero (ignoring the inner sphere for a moment)? $\endgroup$
    – Triatticus
    Commented Aug 21, 2023 at 16:18
  • $\begingroup$ oh i think i understand it, thanks! $\endgroup$
    – Helpme
    Commented Aug 22, 2023 at 3:30

1 Answer 1

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The question is isn't this exactly the same as the electric field of an isolated sphere, where the outer shell doesn't exist?

Yes, that is correct.

Does the negatively charged outer shell not add to the electric field? Why is this the case?

It does not. A spherical shell of charge divides the space into two regions, an inside region and an outside region. In the outside region the field it produces is the same as a point charge at the center. In the inside region the field it produces is zero. This is proven in Newton's shell theorem

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