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MAgnitude of electric Field

  1. Apr 2, 2008 #1
    1. The problem statement, all variables and given/known data
    A solid sphere of radius 40.0 cm has a total positive charge of 46.0 µC uniformly distributed throughout its volume. Calculate the magnitude of the electric field:

    10.0 cm from the center of the sphere
    40.0 cm from the center of the sphere
    60.0 cm from the center of the sphere



    2. Relevant equations

    E=kq/r^2



    3. The attempt at a solution

    8.99e9*46e6/60cm=6.892333333e15
     
  2. jcsd
  3. Apr 2, 2008 #2
    Let's call the charge enclosed by a sphere of radius r (with r < R) Q', and the charge of the entire sphere, Q.

    The ratio of the charges is thus [tex]\frac{Q'}{Q}=\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3}[/tex]

    or:

    [tex]Q'=Q\frac{r^3}{R^3}[/tex]

    Using this, and the fact that [tex]\int \vec{E}.\vec{da}=\frac{q}{\epsilon _0}[/tex] from Gauss's Law.

    You can calculate the electric field at any point within the sphere.
     
    Last edited: Apr 2, 2008
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