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Gauss' law for point charge inside sphere off center

  1. Jun 3, 2015 #1
    1. The problem statement, all variables and given/known data
    If a point charge is inside a Gaussian sphere but is off center, why is its electric field still Qenc/(e0*4*pi*r^2)?


    2. Relevant equations
    surface integral of E*da=Qenc/e0


    3. The attempt at a solution
    If we draw cones out from the charge. the 2 surfaces from the cones' intersection has a charge/area. The bottom of the cone (circle) has areas A1 and A2. The slant of each cone is r1 and r2.

    Then, the teacher said (charge density)(A1/r1^2)=(charge density)(A2/r2^2). I don't know how he jumped to this. He said this proves that the electric field is still the above.
     
  2. jcsd
  3. Jun 3, 2015 #2

    Dick

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    The electric field is not the same. It's stronger where the charge is closer to the surface. What is the same is the total electric flux (the integral of the field through the surface). You teacher is trying to give you an arguement to show Gauss's Law is true. The field gets stronger over some region but it's compensated by getting proportionately weaker over another region but in such a way that the total sum is constant.
     
    Last edited: Jun 3, 2015
  4. Jun 8, 2015 #3
    =So for A1, let's say that the charge is closer to that than A2. Because E1=(charge density *A1)/R1^2 and E2=(charge density *A2/R2^2). We can say that magnitude E1=magnitude E2? So for A1, even though the charge is closer to that area, A1<A2. For A2, even though the charge is farther away from that, the fact that A2>A1 compensates for that?
     
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