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Find the electric field strength inside a solid sphere.

  1. Oct 1, 2008 #1

    _F_

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    1. The problem statement, all variables and given/known data
    A solid sphere 25cm in radius carries 14uC, distributed uniformly throughout its volume. Find the electric field strength (a.) 15cm, (b.) 25cm, (c.) 50cm from the sphere's center.

    R = .25m
    Q = 14 * 10^-6 C

    2. Relevant equations
    p = q/(volume)
    E = (kq)/r^2


    3. The attempt at a solution
    a.) I have absolutely no idea where to even begin here. This is what I really need help with. I'm not sure if the answer relates to the other two distances, so i'll show how I got those. For some reason, E = (kq)/r^2 fails here.

    b.) E = (kq)/r^2 = (9.0*10^9 Nm^2/C^2)(14*10^-6 C)/(.25m)^2 = 2.02 * 10^-6 N/C

    c.) E = (kq)/r^2 = (9.0*10^9 Nm^2/C^2)(14*10^-6 C)/(.50m)^2 = 504 * 10^3 N/C
     
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  3. Oct 1, 2008 #2

    gabbagabbahey

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    The field inside a uniformly charged sphere is not kq/r^2. To determine the actual relationship, the easiest way is to use Gauss's Law. Are you familiar with Gauss's law?
     
  4. Oct 1, 2008 #3

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    Yes I am familiar with Gauss's Law.

    flux = integral(EAcos(theta)) = q_enclosed/e_0.

    I'm still not sure where to go from here, though.
     
  5. Oct 1, 2008 #4

    gabbagabbahey

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    What is [itex]Q_{enclosed}[/itex] at r= 15cm?
     
  6. Oct 1, 2008 #5

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    I know its not 14uC. It must be something else, but I can't seem to figure out what.

    Q = (e_0)*E*A ?

    A = 4(pi)r^2 = 4(pi)(.15m)^2 = .2827m^2 ?
     
  7. Oct 1, 2008 #6

    gabbagabbahey

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    Well the 14uC is distributed uniformly throughout the sphere, so what is the charge density of the sphere then?
     
  8. Oct 1, 2008 #7

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    I would say Q/A = 14*10^-6 C * .2827 m^2 = 5*10^-5 C/m^2

    But plugging that into: E = (5*10^-5 C/m^2)/e_0 = 5.59*10^6 N/C.

    But my book says E = 1.21 MN/C
     
    Last edited: Oct 2, 2008
  9. Oct 2, 2008 #8

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    Can anyone point me in the right direction here?

    I still can't wrap my head around this. :(


    If the charge density is p = q/volume. Then p = (14*10^-6 C)/(4/3*pi*(.25m)^2 = 2.14*10^-4 C/m^3

    But how do I use this?
     
  10. Oct 2, 2008 #9

    gabbagabbahey

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    Well, [itex]Q_{enclosed}[/itex] is the charge enclosed by your Gaussian surface. For part (a), the Gaussian surface does not enclose the entire volume; and so it does not enclose the entire charge. What volume is enclosed for r=15cm? Shouldn't the charge enclosed be that volume times the density that you just calculated?
     
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