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What is the electric field due to hollow sphere at R=z?

  1. Dec 31, 2015 #1
    So I derived the E-field of a hollow sphere with a surface charge σ at z and I got:

    [tex]E(r)=\hat{z}\frac{\sigma R^2}{2\varepsilon _{0}z^2}\left ( \frac{R+z}{\left | R+z \right |}-\frac{R-z}{\left | R-z \right |} \right )[/tex]

    at z>R, the equation becomes:

    [tex]E(r)=\hat{z}\frac{\sigma R^2}{\varepsilon _{0}z^2}[/tex]

    then at z<R:

    [tex]E(r)=0[/tex]

    as expected.

    However, the equation would explode at z=R, since the denominator of the second term in the right hand side equation becomes zero. Now, how do I get over this? and get the E-field at z=R. Any alternate solution to overcome the 0/0?
     
    Last edited by a moderator: Dec 31, 2015
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  3. Dec 31, 2015 #2

    Orodruin

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    The field has a well defined finite limit from both sides. It is discontinuous at z = R, but that is expected from the surface charge.
     
  4. Dec 31, 2015 #3

    jtbell

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    A surface charge density with zero thickness is an idealization. In the real world, a surface charge always has some small thickness, and the field makes a rapid but continuous transition.
     
  5. Dec 31, 2015 #4

    Orodruin

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    Also, in the idealised case, asking for the electric field at the exact point of the surface is not that different from asking about the field at the exact point of a point charge. It is not really surprising that it does not have a particular value and is discontinuous.
     
  6. Dec 31, 2015 #5
    Thanks for all valuable input! Just a follow up, how do we explain the case where there is a Potential V at z=R but E is discontinuous?
     
  7. Dec 31, 2015 #6

    Orodruin

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    This is no stranger than any continuous function with discontinuous derivatives, such as ##|x|##.
     
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