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Generating Expressions For Electric Flux

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle with charge Q is located immediately above the center of the flat face of a hemisphere of radius R as shown in the figure below.

    (a) What is the electric flux through the curved surface? (Use any variable or symbol stated above along with the following as necessary: ε0.)

    (b) What is the electric flux through the flat face? (Use any variable or symbol stated above along with the following as necessary: ε0.)


    2. Relevant equations



    3. The attempt at a solution
    Taking the surface area equation for a sphere, and dividing by two: [itex]2 \pi r^2[/itex]

    Area equation for a circle: [itex]\pi r^2[/itex]

    The expression for (a): [itex]\Phi = E2 \pi r^2[/itex]

    The expression for (b): [itex]\Phi = E \pi r^2[/itex]

    These are incorrect. What did I do wrong?
     

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    Last edited: Feb 9, 2013
  2. jcsd
  3. Feb 9, 2013 #2

    TSny

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    You didn't express the answers in terms of the symbols required. Part (b) doesn't look correct.
     
  4. Feb 9, 2013 #3
    Well, what other relationships should I use, in order to put it in the form that question asks for?
     
  5. Feb 9, 2013 #4

    TSny

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    What does E stand for in your answers? What produces E? Can you express E in terms of the symbols Q, R, and εo?
     
  6. Feb 9, 2013 #5
    Oh, so for (a): [itex]\Phi = \large \frac{Q}{2 \epsilon_0} [/itex] Which would make sense, seeing as if the object were a sphere, and it enclosed the charge at its center, the whole sphere would experience an electric flux of [itex]\Phi = \large \frac{Q}{\epsilon_0}[/itex]; but since it is half a sphere, half of the of electric field is emanating away from the bottom half of this sphere.
     
  7. Feb 9, 2013 #6

    TSny

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    Yes, that looks correct.
     
  8. Feb 9, 2013 #7
    You said that my answer for part (b) didn't look correct, care to elaborate? I'd appreciate it.
     
  9. Feb 9, 2013 #8

    TSny

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    In general the flux through an area is not simply the product of the magnitude of E and the area. That only works if the field is uniform over the surface and perpendicular to the surface. Neither of these conditions is met for the flat face.
     
  10. Feb 9, 2013 #9
    The field isn't parallel to the area vector at any point on the circular surface? How am I to solve this part, then?
     
  11. Feb 9, 2013 #10

    TSny

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    Think Gauss' law which applies to any closed surface.
     
  12. Feb 9, 2013 #11
    This isn't a closed surface, though.
     
  13. Feb 9, 2013 #12

    TSny

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    Right. The flat surface by itself is not a closed surface. But is it part of a closed surface?
     
  14. Feb 9, 2013 #13
    Yes. What does that imply?
     
  15. Feb 9, 2013 #14

    TSny

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    What does Gauss' law tell you about the total flux through this particular closed surface?
     
  16. Feb 9, 2013 #15
    Does it have to do with the amount of electric fields going into the surface equals the amount exiting? Meaning the flux is the same for both surfaces? Why is this true?
     
  17. Feb 9, 2013 #16

    TSny

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    Can you state Gauss' law?
     
  18. Feb 9, 2013 #17
    I looked at Wikipedia's definition, but I am unable to state it in my own words.
     
  19. Feb 9, 2013 #18

    TSny

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    Wkipedia says "The electric flux through any closed surface is proportional to the enclosed electric charge". How much charge is enclosed in your closed surface?
     
  20. Feb 9, 2013 #19
    There isn't any charge within the closed surface; only electric fields pass through the surface.
     
    Last edited: Feb 9, 2013
  21. Feb 9, 2013 #20

    TSny

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    That's right. Since Gauss' law states that the total flux through the closed surface is proportional to the charge within, what can you deduce about the value of the total flux?
     
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