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Use Gauss’s Law to prove that the electric field anywhere inside the hollow of a charged spherical shell must be zero.

My attempt:

$$\int \mathbf{E}\cdot \mathbf{dA} = \frac{q_{net}}{e}$$

$$\int E \ dAcos\theta = \frac{q_{net}}{e}$$

$$E \int dA = \frac{q_{net}}{e}$$

$E\ 4\pi r^2 = \frac{q_{net}}{e}$ and since it is a hollow of a charged spherical shell the $q_{net}$ or $q_{in}$ is $0$ so: $E = 0$.

Is my reasoning on this problem correct? Essentially $E$ is $0$ because there is no charge enclosed.

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  • $\begingroup$ Spherical symmetry implies that $\mathbf{E}=E(r)\ \hat{e}_r$, where $\hat{e}_r$ is a unit vector in the radial direction. If your charged shell is conducting, then the assumption of spherical symmetry holds. $\endgroup$
    – suresh
    May 12 '14 at 8:52
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You may have forgotten to consider the case where $\vec E \perp\vec A$. Then, also flux is zero. But, it is easy to tell using symmetry that then $\vec E$ would form closed loops which is not permissible. Hence, $E$ has to be zero units.

And yes, your reasoning is correct. You can show this for any (imaginary) shell inside your shell. Hence, Electric Field is zero everywhere.

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  • $\begingroup$ Does this mean the E field inside a hollow inside both a conductor and dielectric is 0 or just the charge? $\endgroup$ May 12 '14 at 5:02
  • $\begingroup$ As long as you can, by symmetry, say that E though every dA is same. $\endgroup$
    – evil999man
    May 12 '14 at 5:04
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It isn't enough to show that the net field over some randomly chosen surface within the shell is zero. After all, this is also true of a closed surface between the plates of a capacitor, but the field is not zero there.

You need to use the spherical symmetry and choose a sperical shell centred at the centre of your sphere. Then you can use the symmetry to argue that the field must be everywhere normal to your surface and everywhere have the same value.

Apologies if this is already implicit in your working - it isn't clear to me whether you've considered this or not.

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