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Electric Potential of Spheres

  1. Aug 14, 2006 #1
    I have two problems that confuse me for similar reasons. The first one:

    Find the potential [itex]\varphi[/itex] of an uncharged conducting sphere outside of which a point charge [itex]q[/itex] is located at a distance [itex]l[/itex] from the sphere's center.
    The second one:

    A system consists of two concentric conducting spheres, with the inside sphere of radius [itex]a[/itex] carrying a positive charge [itex]q_1[/itex]. What charge [itex]q_2[/itex] has to be deposited on the outside sphere of radius [itex]b[/itex] in order to reduce the potential of the inside sphere to zero?​
    There's more to the second problem, but this first part confused me enough.

    For the first one, I originally went about it by defining the inner sphere's radius as [itex]R[/itex], and then using the law of cosines to find the distance between the point charge and the surface of the sphere as a function of [tex]\theta[/itex]. This, however, ended in failure, with undefined results.

    Then I thought, "maybe the potential of the sphere is located at the center of the sphere?" So I wrote down [tex]\varphi_0 = \varphi_q + \varphi_s[/tex]. Since [tex]\varphi_q[/tex] is [tex]\frac{q}{4\pi \epsilon_0 l}[/tex] if the potential is at the center of the sphere, and [tex]\varphi_s[/tex] is [tex]\frac{0}{4\pi \epsilon_0 R}[/tex], the potential of the sphere must be [tex]\frac{q}{4\pi \epsilon_0 l}[/tex]. This is the right answer, but I am still confused - I thought the potential of a sphere should be treated as if it were on the sphere's surface, not as if it were at the center?

    For the second one, I still wasn't sure if potential should be treated as surface or center, so I calculated blindly.

    If the potentials are at the center, as gave me the correct answer for the first problem, then in order for the potential of the center sphere to become zero the potential of the outside sphere must cancel it out.

    [tex]\varphi_a = -\varphi_b[/tex]

    After integrating those expressions from their differential parts, I concluded that [tex]\frac{-q_1 a}{b} = q_2[/tex], which is the wrong answer. I then tried it with Gauss's Law, but I still got the wrong answer.

    I'm probably messing up because I don't understand electric potential as it applies to spheres.

    In summary, I have two main questions:
    1. Did I do the first problem correctly, and, if so, why is it correct?
    2. How am I supposed to set up the second problem?

    Thank you for your time.
  2. jcsd
  3. Aug 16, 2006 #2
    I've figured out the first question, so I don't need help with that one, but the second one (with the two spheres) still confuses me.
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