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

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1. Dec 31, 2015

### 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?

Last edited by a moderator: Dec 31, 2015
2. Dec 31, 2015

### Orodruin

Staff Emeritus
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.

3. Dec 31, 2015

### Staff: Mentor

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.

4. Dec 31, 2015

### Orodruin

Staff Emeritus
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.

5. Dec 31, 2015

### 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?

6. Dec 31, 2015

### Orodruin

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