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Electric field strength (should be easy)

  1. Feb 2, 2005 #1
    What is the electric field strength in a region where the flux through a: [tex]1.0cm \times 1.0cm[/tex] flat surface is: [tex]65N\frac{m^2}{C}[/tex], if the field is uniform and the surface is at right angles to the field?

    []
    []----> E
    []

    Since the field points at a right angle to the surface we have a norm of 1. (cos(90) = 1).

    So we have:

    [tex]
    \phi = 65N\frac{m^2}{C}
    [/tex]
    [tex]
    A = (1.0cm)^2 = (0.01m)^2 = 0.0001m^2
    [/tex]
    [tex]
    E= unknown
    [/tex]

    Therefore:

    [tex]
    \phi = E*A
    [/tex]

    [tex]
    65N\frac{m^2}{C}=E (0.0001)
    [/tex]
    [tex]
    E=\frac{65}{0.0001}\frac{Nm^2}{C}
    [/tex]

    Thus:
    [tex]
    E=650000N\frac{m^2}{C}
    [/tex]

    Did I do this math correctly? I just feel like I am doing something wrong here. Thanks
     
  2. jcsd
  3. Feb 2, 2005 #2
    Your answer is correct, but let me just point something out.
    The norm is a unit vector that is perpendicular to the surface, so in this case its parallel to the electric field so the angle (theta) between the norm and the E is 0. cos(0) = 1,
    cos(90) = 0 ( I think you knew this but just made careless/typographical error )
     
  4. Feb 2, 2005 #3
    hehe, so careful with latex... not so careful with plain text.

    I just started this physics course, so all the reinforcement (even mistakes) is awesome... thanks for the double check :)
     
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