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Homework Help: Can't Find Error in Application of Gauss's Law

  1. Jan 15, 2010 #1
    Can't find the error I made... Rudimentary problem i know but i can't find where my mistake is...

    1. The problem statement, all variables and given/known data
    A nonconducting sphere is made of two layers. The innermost section has a radius of 6.0 cm and a uniform charge density of -5.0 C/m3. The outer layer has a uniform charge density of +8.0 C/m3 and extends from an inner radius of 6.0 cm to an outer radius of 12.0 cm. Determine the electric field for: ... (b) 6.0 cm < r < 12.0 cm


    2. Relevant equations
    Gauss's Law [tex]\oint\vec{E}\text{dA=}\frac{Q_{encl}}{\epsilon_{0}}[/tex]

    3. The attempt at a solution
    Basically I would find the electric field due to the inner layer and then the outer layer and add them together.
    Starting with the outer layer:
    [tex]\tex{E(4}\pi\tex{r^{2})=}\frac{1}{\epsilon_{0}}\tex{(}\frac{r^{3}}{r_{0}^{3}}\tex{)Q}[/tex]
    [tex]\tex{E=}\frac{1}{\tex{4}\pi\epsilon_{0}}\tex{(}\frac{r}{r_{0}^{3}}\tex{)Q}[/tex]
    [tex]\tex{E=}\frac{1}{\tex{4}\pi\epsilon_{0}}\tex{(}\frac{1}{0.12^{3}}\tex{)(}\frac{4}{3}\pi\cdot\tex{0.12^{3})(8)r[/tex]
    [tex]{E=}\frac{8}{3\epsilon_{0}}{r}[/tex]
    [tex]{E=3.0}\cdot{10^{11}r}\hspace{1 pc}\frac{N\cdot\tex{m}}{C}[/tex]
    so far so good i think
    now for the internal part...
    [tex]{E(4}\pi\tex{r^{2})}=}\frac{Q}{\epsilon_{0}}[/tex]
    [tex]{E=}\frac{1}{4\pi\tex{r^{2}}\epsilon_{0}}{(}\frac{4}{3}\pi\tex{0.06^{3})(-5)[/tex]
    [tex]{E=}\frac{-0.00108}{3\cdot\tex{(8.9\cdot{10^{-12})r^{2}}}}[/tex]
    [tex]{E=-4.1}\cdot{10^{7}}\frac{1}{r^{2}}[/tex]

    however there is the problem... i don't see where my mistake was in that second part, but the book says the second part of the Electric Field from the core sphere should be -1.1x10^8/r^2......

    so confused maybe its just too late (early) but any help is appreciated
     
    Last edited: Jan 15, 2010
  2. jcsd
  3. Jan 15, 2010 #2
    Does the book explicitly state that the electric field from the inner sphere is -1.1 x 10^8/r^2, or does it state the electric field between two radial distances?
     
    Last edited: Jan 15, 2010
  4. Jan 15, 2010 #3
    The book gives the answer as exactly:
    -(1.1 x 10^8 N*m^2/C)/r^2 + (3.0 x 10^11 N/C*m)r

    And as my outer shell matches the second part of their answer exactly, i can only assume that the first part is what my inner shell should amount to...
     
  5. Jan 15, 2010 #4

    vela

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    Your calculation for the outer shell assumes you have a solid sphere, including the region [itex]r<0.06~\mbox{m}[/itex]. To compensate for that, the book used a charge density of -13 C/m^3 for the inner sphere.
     
  6. Jan 15, 2010 #5
    OH! I see!!

    Yea for anyone else that might stumble here from google like i always do, I forgot to subtract out the inside of the outer sphere.

    Thank you vela!!
     
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