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Gauss's Law and charge distribution

  1. Jan 24, 2009 #1
    1. The problem statement, all variables and given/known data

    A spherically symmetric charge distribution produces the electric field [tex]E = (5000r^2) N/C[/tex], where r is in m.

    a) What is the electric field strength at r = 20cm?

    b) What is the electric flux through a 40cm diameter spherical surface that is concentric with the charge distribution?

    c) How much charge is inside this 40cm diameter spherical surface?

    2. Relevant equations

    [tex]E = (5000r^2) N/C[/tex]

    3. The attempt at a solution

    I am not sure if I did this right?

    a)

    [tex]E = (5000r^2) N/C[/tex]

    [tex]E = (5000(0.2m)^2) N/C[/tex]

    = [tex]200 Nm^2/C[/tex]

    b)

    40cm = diameter
    20cm = radius

    [tex]E = (5000(0.2m)^2) N/C[/tex]

    = [tex]200 Nm^2/C[/tex]

    c)

    [tex]\phi_{e}=\frac{q}{\epsilon_0}[/tex]

    [tex]q=(\phi_{e})(\epsilon_0)[/tex]

    [tex]q=(200 Nm^2/C)(8.85x10^{-12} C^2/Nm^2)[/tex]

    [tex]= 1x10^{-9} C[/tex]
     
  2. jcsd
  3. Jan 24, 2009 #2

    Dick

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    You are stumbling on the units. They should have written E=5000*(r/m)^2*N/C, in my opinion. I think that's what they really mean by "where r is in m". Now if you do a) you get E=5000*(0.2m/m)^2*N/Q. Now you have correct E field units of N/C. For the flux in b) you need to integrate that field over the surface of the sphere. Now you pick up the extra m^2 in the units. But you have to multiply the E field from a) by the area of the sphere.
     
  4. Jan 24, 2009 #3
    Ok, I redid it:

    a)

    [tex]E = (5000r^2) N/C[/tex]

    [tex]E = (5000(0.2m/m)^2) N/C[/tex]

    = [tex]200 N/C[/tex]

    b)

    40cm = diameter
    20cm = radius

    [tex]\phi_{e}=\oint EdA = EA_{sphere}[/tex]

    = [tex](200 N/C)(4\pi(0.2m)^2)[/tex]

    =[tex]101 Nm^2/C[/tex]

    c)

    [tex]\phi_{e}=\frac{q}{\epsilon_0}[/tex]

    [tex]q=(\phi_{e})(\epsilon_0)[/tex]

    [tex]q=(101 Nm^2/C)(8.85x10^{-12} C^2/Nm^2)[/tex]

    [tex]= 8.9x10^{-10} C[/tex]
     
  5. Jan 24, 2009 #4

    Dick

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    That looks much better.
     
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