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Homework Help: Coulumb's law: What is the net charge of the shell?

  1. Aug 23, 2009 #1
    1. The problem statement, all variables and given/known data
    A nonconduction spherical shell, with an inner radius of 4.0 cm and an outer radius of 6.0 cm, has charge spread nonuniformly through its volume between its inner and outer surfaces. The volume charge density p is the charge per unit volume, with the unit coulumb per cubic meter. For this shell p = b/r, r is the distance in meters from the center of the shell and b = 3.0µ C/ m^2. What is the net charge of the shell?

    2. Relevant equations

    Density = mass / volume

    or p = mass / volume

    volume of a sphere (3/4) * pi * r^2

    Coulumb's law: F= k * (Q1 * Q2)/r^2

    F = m*a

    K= 8.99 * 10^9 N* m^2/C^2

    q = n * e

    e = 1.602 * 10^-19 C

    3. The attempt at a solution
    I'm confused. It gives me two radius's. So how do I start this problem. Hint.
  2. jcsd
  3. Aug 23, 2009 #2
    You're supposed to find the total charge, and you're given the density function p, so go integrate!
  4. Aug 23, 2009 #3
    What do you mean by integrate, do you mean I shall take the derivative.
  5. Aug 23, 2009 #4


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    None of these equations are relevant here. Instead of listing every equation from your formula sheet, you might try looking up the term "charge density" since you don't seem to understand what it means.

    Using this definition, if I gave you an infinitesimally small piece of material, with charge density [itex]\rho[/itex] and volume [itex]dV[/itex], could you tell me the amount of charge [itex]dq[/itex] enclosed in that volume?

    You should, of course, be able to say immediately that the amount of charge is given by [itex]dq=\rho dV[/itex].

    Your shell is made up of a whole bunch of little pieces of charge just like this, and so to find the total charge enclosed in its volume so simply add them all up, or integrate:

    [tex]Q=\int\rho dV[/tex]
  6. Aug 23, 2009 #5
    Ok, so I have to integrate it. Yea I'm barely learning this stuff, and it's confusing when you apply calculus to physics.
  7. Aug 23, 2009 #6
    So it becomes Q= p * v

    Because p is constant and the antiderivative of DV is V.
  8. Aug 23, 2009 #7


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    Ermmm...*cough* *cough*...

    That doesn't sound much to me like [itex]\rho[/itex] is constant in this case:wink:
  9. Aug 23, 2009 #8
    So it becomes Q = (p^2) /2 * V
  10. Aug 23, 2009 #9


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    No, the integral [itex]V\int\rho d\rho[/itex] would equal that....but that's not what you are trying to calculate.

    You need to express the volume element [itex]dV[/itex] in spherical coordinates....do you know how to do that?
  11. Aug 23, 2009 #10
    Yea I do, but I'm very rusty on it. Ouch.
  12. Aug 23, 2009 #11


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    Try a little elbow grease to get the rust off then.:wink:
  13. Aug 23, 2009 #12
    Thanks man for the help.
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