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Homework Help: Electric flux of surface

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data
    A flate surface with area 0.14m^2 lies in x-y plane in a uniform electric field given by E=5.1i+2.1j+3.5k kN/c

    find the flux through this surface


    2. Relevant equations

    I know that flux= EA which E and A are given so I multiply The area with the victor E but I got a wrong answer

    3. The attempt at a solution
    0.14m^2(5.1i+2.1j+3.5k kN/c)=7.14e-1i+2.94e-1j+4.9e-1z

    but the correct answer is 490Nm^2/C,

    how come can someone please help me. thanks.




    1. The problem statement, all variables and given/known data
    3)A 10-{ nC} point charge is located at the center of a thin spherical shell of radius 8.0 { cm} carrying -20 {{ nC}} distributed uniformly over its surface.

    A.What is the magnitude of the electric field 2.0 {cm} from the point charge and the direction of the field?

    B. What is the magnitude of the electric field 15 { cm} from the point charge?

    2. Relevant equations
    E outside spherical =Q/4pi[tex]k\ =\ 1.3806503(24)\ \times\ 10^{-23}\ J\ K^{-1}[/tex]r^2

    and E inside= Qr/4pi[tex]k\ =\ 1.3806503(24)\ \times\ 10^{-23}\ J\ K^{-1}[/tex]r^3


    3. The attempt at a solutionR^3
    A)and since 8.0cm>2.0cm it's outside so I used the outside equation
    (10e19)(2e-2)/(4pi(9e-2)(8e2^3)=3.45e3

    but the correct answer is 230kN/C

    b) since 15cm>8.0cm the E is inside therefore
    (.08)/4pi(9e-12)(.15)=4.19.e3

    but my answer is wrong

    1. The problem statement, all variables and given/known data
    2)A sock comes out of the dryer with 10^12 excess electron.
    What's the electric flux through a surface surrounding the socks?

    2. Relevant equations
    Flux= EA
    but since the area was not given
    flux= Qenclosed?[tex]k\ =\ 1.3806503(24)\ \times\ 10^{-23}\ J\ K^{-1}[/tex]

    3. The attempt at a solution
    10^12.9e-12=1.e23

    but correct answer is 1.18e4Nm^2/C

    please help me because I have a test on flux tomorrow and without understanding this might cause me. thanks.
     
  2. jcsd
  3. Feb 3, 2009 #2

    tiny-tim

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    Homework Helper

    Welcome to PF!

    Hi katta002! Welcome to PF! :smile:
    You need, not the flux through the area, but the component of flux in the direction normal (perpendicular) to the area.

    So the normal to the area is … ?, and the component of flux in that direction is … ? :smile:

    (and how are you doing on the others?)​
     
  4. Feb 3, 2009 #3
    Re: Welcome to PF!

    Thank you for welcoming me PF and boy am i glad to find this website
    So I multiply the area by the i direction, j direction and K direction and find the sum of all direction but still didn't get the answer
    i=.714
    j=.294
    k=.49
    the sum is 1.498, I still don't get it.

    and the rest that I posted I'm still facing difficulties so could u break it down for me, thanks and God bless you
     
  5. Feb 3, 2009 #4

    tiny-tim

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    that's not the component in the normal direction …

    it's not even a dot-product of anything.

    What is the normal direction (of the area)?​
     
  6. Feb 3, 2009 #5
    pi*r^2
     
  7. Feb 4, 2009 #6

    tiny-tim

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    uhh? :confused:

    The plate lies in the x-y plane …

    so the normal direction is the z-direction (k) …

    so the question is asking what is the component of flux in the k-direction?

    (btw, did you find out how to post in the academic guidance forum?)
     
    Last edited: Feb 4, 2009
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