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Homework Help: Calculate the flux of the indicated electric field vector through the circle

  1. Sep 20, 2009 #1
    Calculate the flux of the indicated electric field vector through the surface. (E = 130, = 66.0°.)

    I know the differential equation for flux d[tex]\phi[/tex]= A [tex]\bullet[/tex] d S

    My textbook told me that "if there is zero total charge within the closed surface S, there is no net flux of the electric field vector through S."
    I guessed that the total flux was 0, but that was wrong...
    Or maybe it's
    [tex]\int[/tex] (from O to .05m) <130,0> [tex]\bullet[/tex] [tex]\pi[/tex] r ^2(<cos60,sin60>)
    But I'm not really even sure how to do an integral with a dot product(would you dot them first?), and I'm fairly certain that's wrong...

    Any ideas?
    Last edited: Sep 20, 2009
  2. jcsd
  3. Sep 20, 2009 #2


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    Welcome to PF!

    Hi cwatki14! Welcome to PF! :smile:

    (type \cdot, not \bullet :wink:)
    A "closed surface" mans a surface with no boundary (like a sphere), but S does have a boundary … it's the circle round S :wink:

    (I suspect you're confusing this with a topologically closed (as opposed to open) surface … a surface which includes its boundary)
  4. Sep 20, 2009 #3
    So is the flux then
    and then you take the integral to get
    and I'm not sure when to go from here...
    Last edited: Sep 20, 2009
  5. Sep 22, 2009 #4
    So I was pretty sure I had it when I realized you can just take the dot product between the area and the electric field.

    Therefore I was led to this answer.
    A dot E= 130 * pi(.05)^2 * cos 66
    Which is equal to the magnitude of the electric field multiplied by the magnitude of the area multiplied by the angle in between the two vectors (reference the pic above). However, my online homework program quickly told me I was wrong! Blergh!!! Any suggestions?
  6. Sep 22, 2009 #5


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    Hi cwatki14! :smile:

    (have a pi: π and a dot: · and try using the X2 tag just above the Reply box :wink:)
    That's right … you only need to integrate if the field isn't constant, or if the surface (or its boundary) isn't planar. :smile:
    Looks ok to me :confused:

    (Are you sure it isn't 150, not 130?)
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