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Suppose we take spherical conductor which is having both positive and negative charges but as a whole is electrically uncharged and is not under the influence of any external Electric field, We can say that the E outside the conductor is zero by gauss law if take the gaussian surface outside the solid conductor as q enclosed by the gaussian surface is zero. But if I take the gaussian surface inside the conductor which is centered around the center of the conductor the q enclosed by that gaussian surface would still be zero as the enclosed volume still has equal positive and negative charges and hence the net enclosed charge is zero and hence E (electric field) inside the conductor is zero. If the conductor is composed of only positive or negative charges then there will be E inside the conductor as q enclosed in this case by a gaussian surface would not be zero as the conductor contains only 1 type of charge. the value of this electric field can easily be calculated by using the gauss law.

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No, the E-field within the conductor, and by extension the E-field on the Gaussian surface, will be zero, regardless of the net charge on the conductor. This is true as long as the radius of the Gaussian surface is less than that of the conductor.

If the conductor carries a net charge, this charge will flow to the surface of the conductor. This is what conductors do: charge is free to move around inside the conductor. The charge collects at the outer surface because the conductor must have the same electrostatic potential everywhere, and hence the E-field inside must be zero.

The charge will affect the E-field outside the conductor. A positive net charge will cause an outward E-field and a negative charge an inward E-field.

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  • $\begingroup$ Ok but is my reasoning in the case of the neutral conductor correct $\endgroup$
    – p0803
    Commented Jul 7, 2019 at 20:21
  • $\begingroup$ Yes, as I said the Gaussian surface will enclose no net charge, regardless of whether the conductor is charged or not, so the E-field on it will be zero. $\endgroup$
    – Puk
    Commented Jul 7, 2019 at 20:35

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