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Find the enclosed charge

  1. Jul 28, 2013 #1
    1. The problem statement, all variables and given/known data
    I'm to find the enclosed charge of the closed surface in the image (not sure how to describe the shape).

    3. The attempt at a solution
    Let [itex]a=4cm[/itex] and [itex]b=5cm[/itex]. The total flux in is then [itex]abE_1[/itex] and flux out is [itex]abE_2[/itex]. Applying Gauss' law then gives the total enclosed charge as [itex]Q=ab\epsilon_0(E_2-E_1)[/itex].

    The step I'm a little unsure about is taking the total flux out to be [itex]E_2[/itex] times the area [itex]ab[/itex], since the electric field [itex]E_2[/itex] is through a slanted surface, not perpendicular. On the other hand, the total flux through that slope would have eventually to pass through an area of [itex]ab[/itex]...
     

    Attached Files:

  2. jcsd
  3. Jul 28, 2013 #2

    gneill

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    Staff: Mentor

    The area [itex]ab[/itex] is correct. You want to sum the component of the flux which is normal to the surface it passes through. If you do the geometry you'll find that for a flat surface at an angle to the flux, this amounts to the cross sectional area which is perpendicular to the field vectors.
     
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