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Electrostatics - Negative charge in sphere

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data

    a metal grounded sphere of radius R in vacuum has a negative charge inside at a distance a from the centre. a) Draw the e-field in the sphere, b) find the force on the charge, c) and determine the maximal and minimal surface charge density, d) as well as the magnitude of the charge on the in,- and outside of the sphere, when we change the surface potential from zero to V.

    3. The attempt at a solution

    I did a-b-c, and I don't know d yet.

    a-b:

    I was thinking Image charges here.

    if I put a charge [tex] q' = q \frac{b}{R} [/tex] at distance [tex] b=\frac{R^{2}}{a} [/tex] from the centre in a straight line from the centre and charge q, the potential is zero on the surface, just like in the problem. the first uniqueness theorem now tells me that the potential inside the sphere is uniquely determined.

    The second uniquess theorem tells me the E-field is uniquely determined, because the total charge on the sphere is given (=0).

    The force is now just a matter of putting symbols in formulas.

    c: the surface charge is minimal/maximal on the places of the sphere closest and furthest from the image charge. So [tex]\nabla E = \frac{\rho}{e_0} [/tex] and you compute [tex]\rho[/tex]

    But what is the total charge when the potential is changed?
     
    Last edited: Feb 28, 2008
  2. jcsd
  3. Feb 28, 2008 #2

    pam

    User Avatar

    "because the total charge on the sphere is given (=0)."

    When the sphere is grounded, the total surface charge on the inner surface of the spherical shell must be the negative of the charge q. This is because V=0, means E=0 just outside. By Gauss's law the net charge must be zero.
    If the sphere's surface is at V, E just outside equals V/R.
    Gauss's law will then give the charge on the outer surface of the sphere.
     
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