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Voltage Query

  1. May 6, 2008 #1

    TFM

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    [SOLVED] Voltage Query

    1. The problem statement, all variables and given/known data

    An insulating spherical shell with inner radius 25.0 cm and outer radius 60.0 cm carries a charge of + 150.0 [tex]\mu C[/tex] uniformly distributed over its outer surface. Point a is at the center of the shell, point b is on the inner surface and point c is on the outer surface

    What will a voltmeter read if it is connected between c and infinity?


    2. Relevant equations

    [tex] V = \frac{1}{4\pi \epsilon _0}\frac{q}{r} [/tex]

    3. The attempt at a solution

    In my text book,it says that any voltage between a point and infinity will equal 0, but entering 0 says it is wrong.

    Using the above equation, how do you put the radius shou.d equal infinty

    Any Ideas?

    TFM
     
  2. jcsd
  3. May 6, 2008 #2

    Shooting Star

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    How have you used the "relevant eqn" here? That is the formula for V at a distance r from a point charge q.

    When a voltmeter is connected between two points, it shows the potential difference between these two points. The potential at infinity is taken to be zero. You have to find the potential at the surface of the sphere, which is non-zero.

    Do you know what is the potential due to the charge inside and outside the sphere? Remember, the charge distribution is spherically symmetric. It must be there in your book or notes.
     
  4. May 6, 2008 #3

    TFM

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    Does this look right, then:

    [tex] \frac{1}{4\pi \epsilion_0}\frac{150*10^{-6}}{c} [/tex]

    ?

    TFM
     
  5. May 6, 2008 #4

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    (What is 'c' doing in the formula? It's just a point on the surface.)

    I don't want you to blindly fiddle around with the formulae and put in values. Neither do I want to give you the answer directly.

    Let's see if you know this. What is the potential at the surface of and inside a spherical conductor with charge q? I know that in the problem it's an insulator, not a conductor, but answer if you can.
     
  6. May 6, 2008 #5

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    I have a diagram for the potential, V of a positively charged spherical conductor, and it has a graph which looks slightly like a trapezium, but the diagonal edges are curved. it gives the voltage inside the sphere to be:

    [tex] V = \frac{1}{4 \pi \epsilon _0} \frac{q}{R} [/tex]

    but it shows alos that it is a constant potential through the sphere

    TFM
     
  7. May 6, 2008 #6

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    You must know that in a spherical conductor the charge distributes itself uniformly only over the surface of the sphere. The potential inside and on the surface of a conductor is the same everywhere. This shows that whenever you have a uniform charge distribution over the surface of a sphere of radius R, the potential inside and on the surface is the same, and is given by:

    [tex]
    V = \frac{1}{4 \pi \epsilon _0} \frac{q}{R}.
    [/tex]

    So, your answer is V-0 = V.

    Outside the sphere, the field and potential is as if the whole charge is concentrated at the centre. This is only true if the charge distribution is spherically symmetric. Even if the whole charge was uniformly distributed over the whole volume of the sphere, outside the sphere the field and potential would be the same. Inside, of course, the potential would not be uniform.

    The graph you have of the potential of the conductor shows that V remains constant from 0 to R, and then falls as 1/r, where r is the distance of a point outside the sphere from the centre, which is the curved portion you were talking about.
     
  8. May 6, 2008 #7

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    I tried entering:

    [tex]8.99*10^{-9}\frac{0.00015}{0.6}[/tex]

    [tex]8.99*10^{-9}\frac{0.00015}{0.6}[/tex]

    Since as it goes out to infininty, where the voltage should be 0, so, as stated, the final potential should be v - 0, but Mastering Physics says this is incorrect.

    I have I missed something?

    TFM
     
    Last edited: May 6, 2008
  9. May 6, 2008 #8

    Shooting Star

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    8.99*10^+9.
     
  10. May 7, 2008 #9

    TFM

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    Thanks for spootting the mistake, silly me :redface:

    The final answer is 2.25 x 10^6 Volts

    Thanks for your assitance, Shooting Star, :smile:

    TFM
     
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