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Homework Help: Gauss's Law and Superposition of Fields

  1. May 19, 2010 #1
    Gauss's Law and Superposition of Fields (edited again, something else wrong)

    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=25861&stc=1&d=1274275158.png

    Right. The shape itself has charge Q, so it has charge density [tex]\frac{Q}{\frac{4}{3} \pi R^3 - \frac{4}{3} \pi (\frac{R}{2})^3} = \frac{6Q}{7\pi R^3}[/tex] Let's call this [tex]\rho[/tex]. If it were filled in entirely, then, it would have charge:

    [tex] Q + \rho V = Q + \frac{6Q}{7\pi R^3}\cdot \frac{4}{3}\pi\left(\frac{R}{2}\right)^3 = \frac{8}{7} Q[/tex]. (V being the volume of the small sphere.)

    By Gauss's Law, the field at the surface of this filled sphere would be:

    [tex] E = \frac{\frac{8}{7}Q}{4 \pi \epsilon_0 R^2} = \frac{2}{7 \pi \epsilon_0 R^2} [/tex]

    Considering now the smaller, negatively charged sphere, this would be carrying a charge of

    [tex]-\rho V = -\frac{6Q}{7\pi R^3} \cdot \frac{4}{3}\pi \left(\frac{R}{2}\right)^3 = -\frac{Q}{7}[/tex].

    The electric field at its surface would then be [tex]\frac{\frac{-Q}{7}}{4 \pi \epsilon (\frac{R}{2})^2} = -\frac{Q}{7 \pi \epsilon _0 R^2}[/tex].

    The electric field at P of the shape, then, is the sum of the individual fields of the filled sphere and the negatively charged sphere - I don't get the answer I'm meant to, and I can't see what I've done wrong. It's probably something really stupid.
     

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    Last edited: May 19, 2010
  2. jcsd
  3. May 19, 2010 #2

    kuruman

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    You have the correct answer. Note that (2/7) + (-1/7) = 1/7, which is the answer you are meant to get.


    Oh, by the way, note that the answer you are meant to get has a 4 in the numerator and in the denominator, so multiply and divide your answer by 4. :wink:
     
  4. May 19, 2010 #3
    Oh yeah. I didn't notice that 4 in the denominator. Knew it was something stupid! Cheers.
     
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