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

[flux!]

So I derived the E-field of a hollow sphere with a surface charge σ at z and I got:

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

at z>R, the equation becomes:

$$E(r)=\hat{z}\frac{\sigma R^2}{\varepsilon _{0}z^2}$$

then at z<R:

$$E(r)=0$$

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?

 

[Orodruin]

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.

 

[jtbell]

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.

 

[Orodruin]

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.

 

[flux!]

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?

 

[Orodruin]

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