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Charged Metallic Balloon (need help understanding a concept)

  1. Jan 28, 2005 #1
    the questioon is this:
    A balloon of radius 34.5 cm is sprayed with a metallic coating so that the surface is conducting. A charge of 1.05 × 10-08 C is placed on the surface. What is the potential on the balloon's surface?

    how i solved it:
    -apply the guass law, which gives me the Electrical Field.

    -apply the formula : Voltage = Ed ,

    where E = electrical field , & d = distance travelled by charges (parrallel to field line)

    n it turn out d = 34.5cm. this is where i don't get. why its d = 34.5cm?

    its it b/c the positive charge move to the center, which travel a distance of 34.5cm? since if the surface is positively charged, it would repel the " + " charge to the centre & attact the " - " charge to surface....right?
  2. jcsd
  3. Jan 28, 2005 #2

    Doc Al

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    Staff: Mentor

    No problem. You should get the field as a function of distance from the center, for points outside of the balloon: [itex]E = kq/r^2[/itex].
    You must integrate from a reference point (usually infinity = 0 potential) to the balloon's surface.

    It's not. It turns out that the potential at a point a distance r from the center of the balloon (where r >= the balloon's radius) will equal kq/r.
  4. Jan 28, 2005 #3

    Andrew Mason

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    Science Advisor
    Homework Helper

    Electrical potential is the work done to bring a unit +charge from infinity to the surface of the sphere, which is:

    [tex]\int_{\infty}^{34.5} E\cdot ds = kQ(\frac{1}{R}-\frac{1}{\infty}) = \frac{1}{4\pi \epsilon_0}(\frac{Q}{34.5}-0)[/tex]

  5. Jan 28, 2005 #4
    ah, ic, thx guy. it seem to clear thing up quite a bit.
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