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I know that this may be a duplicate question but wait, I want to analyse the phenomenon from a different perspective.

For a conducting plane sheet of charge Charged sheet

There should be positive charge at both end so that the electric field at point P1 is cancelled out and thus field inside conductor is zero!


However, in this case Charged Spherical Shell

If we take a Gaussian Surface at S then the inner surface of shell must be -q so that q(inside)=0 and electric field=0

But now, if we take a point at Gaussian surface S and analysis the Net Electric Field it doesn't comes out to be 0. WHY?

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  • $\begingroup$ Can you explain your issue better, please? $\endgroup$
    – silviozzo
    May 4, 2023 at 20:28
  • $\begingroup$ Youve seen that Gauss' Law shows that the field will be zero in the conductor. But how did you arrive at the last part of your question? You haven't shown any work supporting your claim. $\endgroup$
    – Triatticus
    May 4, 2023 at 20:49
  • $\begingroup$ I just found out that my current question resemble a older question asked on stack exchange earlier however it doesn't have satisfactory answer. Can you check that question if my issue seems unclear: physics.stackexchange.com/questions/547484/… $\endgroup$
    – ADITYA DAS
    May 5, 2023 at 6:15
  • $\begingroup$ Which surface is the Gaussian surface $S$? $\endgroup$
    – Farcher
    May 5, 2023 at 7:12
  • $\begingroup$ Yup S is the Gaussian Surface $\endgroup$
    – ADITYA DAS
    May 6, 2023 at 5:24

1 Answer 1

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Taking the Gaussian surface as the dashed line at a distance $r$ from charge $+q$ the electric field at at point on Gaussian surface is $\frac{kq}{r^2}$.

The magnitude of the induced charge $-q$ on the inner surface of the shell must be the same as the magnitude of charge $+q$ and produces an electric field at distances greater than the inner radius of the shell as though it was positioned at the centre of the shell, $\color{red}-\frac{kq}{r^2}$.
Thus the net field inside the conducting shell is zero.

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