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Here is the confusion.

Take a spherical conducting shell of a certain thickness.

When a postive charge is introduced into the shell the inside surface collects negative charges and the outter surface postive charges. So far so good.

Now gauss' law tells be there is no field in the conducting shell and thus its outter and inner surfaces must be equipotential.

However the inside surface has negative charges and thus is at a negative potential and the outter surface has postive charges thus is at a postive potential. And thus the inner and outter surfaces have opposite signs for potential and thus are different.

This seems to falsely contradict the earlier gauss' law conclusion which I know is correct.

Where did I go wrong?

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  • $\begingroup$ Don't forget the potential (or field) due to the charge that you introduced. $\endgroup$
    – garyp
    Commented Mar 22, 2019 at 19:05
  • $\begingroup$ Could you please expand on that statement. I don't seem to understand $\endgroup$
    – VinalV
    Commented Mar 22, 2019 at 19:06
  • $\begingroup$ The inner surface and outer surfaces are equipotential on their own, they obviously dont have the same potential, but the complete outer surface and complete inner surface are equipotentials $\endgroup$
    – Sidharth Giri
    Commented Mar 22, 2019 at 19:19
  • $\begingroup$ Correct but as they have different sign that would imply a potential difference between surfaces and thus an electric field inside the shell? Which of course is false but why $\endgroup$
    – VinalV
    Commented Mar 22, 2019 at 19:21
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    $\begingroup$ @SidharthGiri This is incorrect. The conductor is at equipotential. $\endgroup$
    – my2cts
    Commented Mar 23, 2019 at 10:46

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When a positive charge is introduced into the shell the inside surface collects negative charges and the outer surface positive charges.

These are induced charges and the magnitudes of the positive and negative induced charges are the same.
These induced charges set up electric fields inside the conducting shell which exactly cancel out the electric field produced inside the conducting shell by the positive charge.
Thus there is no potential difference between the inner and outer surfaces of the conductor.

Even though there are negative charges on the inside surface of the conducting sphere the potential of that inside surface would be positive relative to infinity (taken as the zero of potential) as a test positive charge would have to do no work whilst moving though the conducting shell and the outside of the shell is at a positive potential.

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  • $\begingroup$ Thanks. That clears it up $\endgroup$
    – VinalV
    Commented Mar 23, 2019 at 10:46
  • $\begingroup$ @Farcher, charge at the external surface does not induce field inside the shell. Hence there is no charge on the internal surface which would induce a field within the shell. $\endgroup$
    – npojo
    Commented Mar 23, 2019 at 18:33
  • $\begingroup$ @npojo The charge inside the shell produces an electric field which stops at the inner surface. An electric field also appears outside the shell. I used the idea of superposition of the field due to the change inside the shell and the field due to the induced charges to produce no (net) electric field inside the conducting shell. $\endgroup$
    – Farcher
    Commented Mar 23, 2019 at 18:38

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