Spherical conducting shell

  1. An uncharged spherical conducting shell surrounds a charge -q at the center of the shell. The charges on the inner and outer surfaces of the shell are respectively...

    For this question I know that the charges are going to be +q, -q respectively.

    However, what if the center charge wasn't -q but a +q, than would the charges be...

    -q, and +q respectively?

    Simple question, just like to know if I'm thinking in the right direction :+)
  2. jcsd
  3. Yes.
  4. would the electric field at any distance r from the shell then be KQ/r2?
    If so...then when does metallic shielding take place?
  5. It doesn't.

    The easiest way to think of it is by Gauss' Law (for electrostatics, obviously). Take the Gaussian surface to be a sphere outside the conducting sphere. If the charge in the middle is Q, and the charge on the conducting sphere is 0, then the flux ([tex]\Psi[/tex]) through the Gaussian sphere is Q (total enclosed charge). Then, since

    [tex]\Psi = \oint \vec{D} \cdot d \vec{S} = Q \neq 0[/tex]

    Then, D (and E) can't be zero.

    Generally speaking, shields are grounded, though. If this one were grounded, then, while originally neutral, it would take on a charge of -Q once the Q charge is inserted in the sphere, the, by Gauss' Law, the external field is 0.

    I'll let you think about this one, but if you have a region with some field E, then put a floating conducting sphere in it, the field inside the sphere would be 0 and the field elsewhere would be unchanged, so shielding kind of works then.
  6. I am sorry I did not get your point...there is a difference between the floating conducting sphere and an ungrounded metallic shell around a positive charge...the floating conducting sphere does not have postive charge in it like the metallic shell one....
  7. ok, "floating" and "ungrounded" mean the same thing, in case that wasn't clear (sorry). My point was that if you have an electric field, and the source of the field is WITHIN a floating sphere, the sphere will not shield its surroundings (although, it may alter the field if the situation isn't as highly symmetrical as the one in the original post, I'm not sure).

    However, if the source of the field is OUTSIDE the sphere, and the sphere is floating, the region WITHIN the sphere would be shielded from the external field.

    My point was that the conductor does not necessarily need to be grounded for shielding to take place.
    Last edited: May 13, 2008
  8. ohhhh...now I understand.Wow that is interesting. Why does this happen "However, if the source of the field is OUTSIDE the sphere, and the sphere is floating, the region WITHIN the sphere would be shielded from the external field"...is there any specific reason for this to happen?
  9. Look, if you have a conducting spherical shell, and you place a charge Q outside the shell, then the charge Q polarises the shell, i.e. it attracts negative charge towards it, and repels positive charge in the shell. However, the shell remains neutral. The charge on the shell rearranges until the net electric field inside the shell is zero.
  10. To visualize it more easily, imagine a uniform electric field in which an uncharged metal shell is placed.

    Let's say the electric field points from left to right. This means that positive charges (or lack of negative charges, electrons) in the metal shell will start moving to the right and negative charges start moving to the left.
    After a short while, the left of the shell will be negatively charged (excess of electrons) while the right will be positively charged (lack of electrons).
    This will in turn generate it's own electric field, pointing from positive to negative, or right to left.
    The magnitude of this electric field (inside the spherical shell) is exactly the same as the magnitude of the field outside the shell, only in the exact opposite direction. The result is that, within the shell, there is NO net electric field.

    If you know Gauss' law you can easily check this:
    [tex]\oint \vec{E} \cdot d \vec{A} = \frac{Q_{\text{encl}}}{\epsilon_0}[/tex]

    If you take a Gaussian surface (spherical) inside the shell, you are not enclosing any charge and thus the electric field is 0.
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