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My textbook says:

  • Inside a conductor, electric field is zero.
  • The interior of a conductor can have no excess charge in static situation.
  • Electric field just outside a charged conductor is perpendicular to the surface at every point.

I know that these laws are meant for solid conductors or in other words conducting materials alone. But if we had a hollow spherical conducting sphere maybe with air (or any insulator) as the medium inside the sphere, how would these laws be affected if a positive charge ( or any charge for that matter) is placed inside the sphere in air (static condition).

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    $\begingroup$ Have you tried searching the internet? eg google search on your title gives the following as #4 and #5 physics.bu.edu/py106/notes/Conductors.html physicsclassroom.com/class/estatics/Lesson-4/… $\endgroup$
    – sammy gerbil
    Commented Jul 28, 2016 at 15:31
  • $\begingroup$ the 1st one was with respect to solid metallic conductors. However, the second did provide a bit of insight. Thanks $\endgroup$
    – Sharpfawkes
    Commented Jul 28, 2016 at 16:45
  • $\begingroup$ the comments are universally true, for the metal of the conductor $\endgroup$
    – Lelouch
    Commented Jul 28, 2016 at 16:53

2 Answers 2

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The macroscopic electric field inside the 'metal' of the conductor is zero in electrostatic conditions. In a hollow cylinder , if a positive charge is place inside the cavity, the field is non zero inside the cavity.

Again, the interior of a hollow shell can hold the positive charge there, because of the induced charges on the inner wall of the cavity. These charges, nullify the field due to the positive charge inside the 'metal' of the conductor. The last conclusion is always true in electrostatic conditions, when the conductor becomes an equipotential.

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  • $\begingroup$ But then in that case, shouldn't the second assumption be invalid? As a Gaussian surface drawn within the conductor would enclose some net amount of flux? $\endgroup$
    – Sharpfawkes
    Commented Jul 28, 2016 at 14:59
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    $\begingroup$ No, it wont, because that gaussian surface would include your test charge inside the cavity + the charge induced on the INNER surface of the cavity, which gives a total charge enclose = 0 $\endgroup$
    – Lelouch
    Commented Jul 28, 2016 at 15:13
  • $\begingroup$ if there was a negative charge inside the hollow sphere and a positive charge on the surface of the sphere, the positive charges would still remain on the surface? If so, why aren't they attracted by the negative charge? $\endgroup$
    – Sharpfawkes
    Commented Jul 28, 2016 at 16:57
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    $\begingroup$ because the field of the negative charge inside will not penetrate the METAL of the conductor to affects the charges on the outside(shielding). $\endgroup$
    – Lelouch
    Commented Jul 28, 2016 at 17:00
  • $\begingroup$ However, if charges are induced in the inner wall, wouldn't an equal amount of charge be induced on the outer surface of the wall. This would produce an electric field unless the sphere is grounded from the outside which is not what i'm assuming. $\endgroup$
    – Sharpfawkes
    Commented Jul 28, 2016 at 17:40
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As there is a positive charge inside then the inner surface would have an equal magnitude negative charge spread around possibly unequally with more nearer to the interior +ve charge with field lines bending to meet inner surface perpendicularly. An equal positive charge is induced on the outer surface. As the electric field inside the conductor is zero there is kind of no communication between inside and out. So if the body is touched (earthed) the outer surface positive charge flows away leaving an apparently uncharged body. The inner cavity with its +charge and - surface charge and interior field remain unaltered. We could put any charge on outer surface and the inner cavity is unaffected. The outer surface charge distribution depends only on the outer shape and not on the inner shape, and vice versa. Curious.

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