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Homework Help: Little question on the potential in a conducting spherical shell

  1. Mar 4, 2006 #1
    The question is:

    A conducting spherical shell of inner radius b and outer radius c is concentric with a small metal sphere of radius
    a < b. The metal sphere has a positive charge Q. The total charge on the conducting spherical shell is –Q. What is the potential of the metal sphere?

    I thought: The potential of the metal sphere is measured when a < r < b and V= Vmetal sphere+ Vinner radius of shell+Vouter radius of shell

    1) The metal sphere is to be considered as a point charge, therefore [tex]V_s_p_h_e_r_e= \frac{kQ}{r} [/tex]

    2) The inner radius has charge -Q in order to compensate the charge of the metal sphere and if a< r < b then r < the inner radius of the metal sphere (b). Because the rule for a spherical shell is:

    [tex] V= \frac{kQ}{r} [/tex] if r > R
    [tex] V=\frac{kQ}{R} [/tex] if r < R

    with R the radius of the spherical shell and r the distance from its center

    V inner radius of shell= [tex]\frac{kQ}{b} [/tex]

    3) The outer radius has charge 0 because a charge of -Q is induced on the inner sphere, a charge of +Q will be found on the outer spherical shell which already was -Q before the electrostatic equilibrium was established. Therefore Q outer radius= 0 and V outer radius=0

    Summerizing:

    Vtotal= [tex]\frac{kQ}{r} + \frac{kQ}{b} = kQ{\frac{1}{r} - \frac{1}{b}[/tex]

    though my textbook tells me Vtotal= [tex]kQ{\frac{1}{a} - \frac{1}{b}[/tex]

    So who is right and who is wrong?
     
  2. jcsd
  3. Mar 4, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    How can you have an r in your answer? You are asked for the potential at a particular point, r = a.

    (1) What's the potential difference between points r = b & r = a?

    (2) What's the potential difference between r = b & r = c? (Don't forget it's a conducting shell.)

    (3) What's the potential difference between r = c & r = infinity? (The outer surface has zero charge.)
     
  4. Mar 4, 2006 #3
    Thank you I've figured it out now :)
     
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