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Electric Field of a charged sphere with cylindrical gaussian surface

  1. Jun 1, 2014 #1
    So the problem statement is:

    A conducting solid sphere (R = 0.167 m, q = 6.63·10–6 C) is shown in the figure. Using Gauss’s Law and two different Gaussian surfaces, determine the electric field (magnitude and direction) at point A, which is 0.00000100 m outside the conducting sphere. (Hint: One Gaussian surface is a sphere, and the other is a small right cylinder.)

    Now with the spherical gaussian surface i've got no doubt..

    The problem with the cylindrical surface is that the electric field it's not constant ,is it?
    in some parts the "area vector" is parallel to the vector of electric field and in some part there's an angle.

    So i'm not sure if i have to find a function of the electric field with respect to the radius of the sphere or how to get ride of this problem.

    What i've first tried was to "assume the electric field" was constant so integrating dA what i've got:
    since the height of cylinder is h=2r (of the sphere).
    and there's flux all over the cyinder: E(2pi*h+2pi*r^2)=Q/ε...
    E(4pi*r+2pi*r^2)=Q/ε.... and so E=Q/(4pi*r+2pi*r^2)ε... but i'm not sure if a i could assume that...
     
    Last edited: Jun 1, 2014
  2. jcsd
  3. Jun 2, 2014 #2

    BvU

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    Reconsider your choice of small cylinder. Guiding principle: don't look for more difficult things to solve but try to find something even easier. Your clues:
    1. the kind of answer to the first part of the exercise,
    2. conducting sphere,
    3. 0.00000100 m is very, very close to the surface
     
  4. Jun 2, 2014 #3
    i could reconsider however it's something i'm being asked for. :(.

    yep i know the answer must be E=Q/(4pi*r^2)ε.

    and i know also the charge on the sphere it's all over the "surface area" and that the distance from the center to the point would be R+0.00000100 m.

    Once i got the Electric field ussing a cylindrical gaussian surface it's just substitution. the problem is i need but i'm not sure how to reach the same result mathematically using that surface :/
     
  5. Jun 3, 2014 #4

    BvU

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    2. so all the charge is at the surface, and uniformly distributed.
    3. so close to the surface that field lines over there are parallel
    So a really puny cylinder standing upright on an almost flat surface would be a nice idea. floor just under the surface, top 0.00000100 m or thereabouts above....
     
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