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Homework Help: Electric field a spherical surface

  1. Mar 23, 2014 #1
    i want to solve this problem via another approach
    Book Solution:

    my approach:
    Coulomb's law for surface charge:


    as we know the filed point is a fix point and i set the name of h instead of z
    (r is in spherical coordinate and haz in cartesian)


    ar in spherical coordinate is equal to below statement in cartesian



    as we know the intergral of Cos(phi) and Sin(phi) in a total period of phi is equal to zero so the main Integrals can be simplified to the following expression:


    i seprate the above integral to two statement


    the below intergal because of is equal zero



    so the output is that is different between this way and the book solution


    what is my problem? and how can find the electric filed in out and in of spherical surface via this approach?

    Last edited: Mar 23, 2014
  2. jcsd
  3. Mar 23, 2014 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Well, there's the posted solution's hard way and then there's your hard way (don't know what you did wrong).
    Then there's the easy way: Gaussian surface, which you should invoke since the problem does not force you to do it the hard way. Just ignore the hint - completely!
  4. Mar 23, 2014 #3
    I am having trouble understanding your notation. Can you clarify what ##\vec{as}## is? It seems to me that you are trying to write the electric field in vector notation. If so, the vector notation is:
    $$\vec{E}=\frac{q}{4\pi\epsilon \,\,r^3}\,\vec{r}$$
    Notice that its ##r^3## in the denominator.

    And yes, Gauss law is a nice way to solve the problem but I guess the problem requires you to take the harder approach.
  5. Mar 23, 2014 #4
    Hello dear user
    thanks for your response.
    yes gauss law is a shortest way to find the electric filed of this question.but i want to know what is my wrong in this approach that i get different answer?

    $$\vec{E}=\frac{q}{4\pi\epsilon \,\,r^3}\,\vec{r}$$
    yes [itex]\vec(as)[/itex] is equal to [itex]\vec(ar)[/itex].in electrodynamic of david giriffts he set different char to avoid confusion with ar in spherical and cylindrical coordinates.
  6. Mar 27, 2014 #5
    i want to know my approach is wrong or i do some mistake?
    Thank you
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