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Gaussian surface of radius $r$ in a non-conducting charged sphere of radius $R$

Gaussian surface of radius r in a non-conducting charged sphere of radius R

By using gauss law with the gaussian surface depicted above we should get a result as follows:

$$\int{E \cdot ds} = \frac{\sum{q_{in}}}{\epsilon}$$

Here I recognize the electric field is due to all the charges present.

However the surface integral of electric field evaluates to zero due to all the charges present outside the gaussian sphere. And charge inside the gaussian surface here is 0. Hence the above equation gives:

$$0 = 0$$

Which does not give any information.

In books I have seen the integral being equated to zero since the charge inside is 0 which goes against my logic.

I have come across another question on this platform with the same doubt where the answer is given on the basis of symmetry but how does it make sense in a mathematical perspective (strictly according to the equation)?

Any help would be appreciated

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  • $\begingroup$ How did you determine that "the surface integral evaluated to zero due to all the charges present outside the Gaussian sphere"? $\endgroup$
    – The Photon
    May 30, 2021 at 17:35
  • $\begingroup$ I was given the explanation: all field lines that enter the gaussian surface exit thus the net flux through it is zero $\endgroup$
    – Draculin
    May 30, 2021 at 17:37
  • $\begingroup$ That result comes from from Gauss's Law. So if they you have a result that comes from gauss's law and you try to apply gauss's law to it, of course you get no new information. $\endgroup$
    – The Photon
    May 30, 2021 at 17:38
  • $\begingroup$ But that result does not strictly come from gauss law right? Or does it? In either way how does that result fail in giving new information if it's true and valid in this case $\endgroup$
    – Draculin
    May 30, 2021 at 17:46
  • $\begingroup$ Hello! I have taken the liberty to resize your images to improve readability. Feel free to rollback if you wish. Thanks! $\endgroup$
    – Jonas
    May 30, 2021 at 17:53

1 Answer 1

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If you are not only aware of the fact that the net electric field inside the hollow sphere is zero but also know that Q (inside the Gaussian Surface) is equal to zero, why are you plugging the values back into the equation? The equation will always hold true and in this case, you get "0=0".

Nonetheless,

For explanation purposes, I am gonna refer to your statement:-"all field lines that enter the gaussian surface exit thus the net flux through it is zero." $$\int^\ E.ds = {ɸ(net)}$$ since ɸ=0, we can say that $$\int^\ E.ds = {0}$$ Now at this point, we have 3 cases

  1. Either ds=0
  2. or E=0
  3. or E.ds=0 (in case the two vectors become perpendicular somehow)

Note that E is due to the charges present outside as well (however, they have no effect on the net flux)

We see that case 1. cannot be true since that indicates that our Gaussian surface doesn't even exist.

In case 3. we see that E.ds isn't zero for all point as well. (try sketching the ELOF due to the hollow sphere at a particular point)

Thus we are left with the result that E must be zero.

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  • $\begingroup$ Oh I think I got it, there can be no case when electric field lines due to all the small charges on the hollow sphere enter a gaussian surface and exit without intersecting another electric field line, hence we can say no field lines exist inside a gaussian surface in this case which would mean that E = 0 $\endgroup$
    – Draculin
    May 31, 2021 at 7:56
  • $\begingroup$ But in that case, "E.ds is not zero for all point" as you post suggests does not make much sense $\endgroup$
    – Draculin
    May 31, 2021 at 7:59
  • $\begingroup$ I was just trying to get that possibility out of the way. You see sometimes the electric field vector can be perpendicular to the area vector of a differential element making E.ds = 0 which makes the electric flux through the Gaussian surface = 0. ( I hope you realise that there is a dot product ) $\endgroup$
    – Parth
    May 31, 2021 at 13:17

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