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I understand that at static equilibrium, the electric field inside a conductor is zero, so the potential difference is zero between any two points within the conductor. However, the surface of a charged solid spherical conductor has a net charge, and thus a net electric field, unlike the inside of the sphere. So why is the electric potential within the conductor and the surface of the conductor the same?

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The field is actually discontinuous at the surface: the discontinuity in the field is proportional to the surface charge density.

The statement "within the conductor and the surface" is to be understood as meaning within the conductor and a point arbitrary close to the surface but inside this surface. The situation you describe is an idealization as, in real conductors, the charge is concentrated in a small boundary around the surface; the thickness of this boundary depends inversely on the conductivity of the material, and goes to zero in the ideal case of a perfect conductor with conductivity $\sigma\to\infty$.

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  • $\begingroup$ So is it technically wrong to say that an entire charged conducting sphere (including the surface) is an equipotential? My textbook says that, hence my original question. $\endgroup$ – ABCDF Dec 24 '16 at 22:06
  • $\begingroup$ Actually my original answer was incorrect. Only the field is discontinuous, aka the derivative of the potential is discontinuous. The potential itself is continuous. Now... I would always make the statement the $\endgroup$ – ZeroTheHero Dec 25 '16 at 2:37
  • $\begingroup$ emphasizing it is an idealization in the limit if arbitrary large conductivity. Technically correct, but to be understood with a caveat. $\endgroup$ – ZeroTheHero Dec 25 '16 at 2:44
  • $\begingroup$ @ZeroTheHero : I suggest you correct your answer according to your comment (continuity of the potential through the boundary). Also, why so is in your first sentence? Derivative of the potential and electrostatic field are the same thing (up to the sign). $\endgroup$ – user130529 Dec 27 '16 at 18:42
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The fact that there is an electric field outside the conductor only means that the potential will change more the further away you go from the surface. The field can be thought of as a derivative (or more precisely, a gradient) of the potential with respect to distance. The potential will change starting from the potential of the conductor at the surface, and then grow or decrease as you get further away from the conductor.

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