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Calculating E field generated by spherically symmetric charged sphere (Gauss' law?)

  1. Jun 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Compute the electric field generated by a spherically symmetric charged sphere of radius R with charge density of [itex]\rho = kr^{2}[/itex]


    2. Relevant equations
    [itex]\oint _S \vec{E} \cdot \vec{dA} = \frac{Q_{enclosed}}{\epsilon_0}[/itex]

    3. The attempt at a solution
    I know that this question involves the application of Gauss' law but I don't really know how? To be honest I'm a bit sketchy on applying Gauss' law to any question. Any help would really be appreciated.
     
  2. jcsd
  3. Jun 20, 2011 #2

    ideasrule

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    Re: Calculating E field generated by spherically symmetric charged sphere (Gauss' law

    Imagine a Gaussian sphere of radius r, centered on the actual sphere of charge. What would the surface integral of E*dA be, in terms of r? Remember that due to symmetry, the electric field has to be constant for constant r, and must be entirely radial.

    Using integration, can you also find Q_enclosed for this Gaussian sphere?
     
  4. Jun 21, 2011 #3
    Re: Calculating E field generated by spherically symmetric charged sphere (Gauss' law

    OK, by integration I've found the charge enclosed by the sphere to be (4pi*k*r^5)/5, but I'm not really sure where to go from here?

    From
    [itex]\oint _S \vec{E} \cdot \vec{dA} = \frac{Q_{enclosed}}{\epsilon_0}[/itex]
    I can see that I need to divide the charge enclosed by epsilon 0 then equate to the surface integral of E*da, but I'm not really sure how to calculate the surface integral of E*da?
    Thanks
     
  5. Jun 21, 2011 #4
    Re: Calculating E field generated by spherically symmetric charged sphere (Gauss' law

    Ok, I've now been told that the surface integral of E*dA in this case goes to E(4pi*r^2) but I'm still not totally sure why.
     
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