Electric field by continuous charge distributions.

[Buffu]

While reading the book, Electricity and magnetism, the author says that electric field just outside a spherical shell is $4\pi \sigma$, on it $4\pi\sigma r_0^2$ ,inside is $0$ and outside is $Q/R^2$.

My derivations :-

For inside,

$E\Delta S = 4\pi Q = 0$ since $Q = 0$.

For outside,

$E \Delta S = 4 \pi {Q} \iff E \cdot 4 \pi R^2= 4 \pi {Q}$

For just outside,

$E \cdot (4\pi r_0^2) = 4\pi \sigma (4\pi r_0^2)$.

But I am unable to derive the one for "on it", any suggestions what should I do ?

So far I get,

$E \cdot (\Delta S) = 4\pi (\sigma 4\pi r_0^2)$.

What should be $\Delta S$ ?

Note for future readers :-

The charge is $4\pi\sigma r_0^2$ not the Field. I read it incorrectly.

 

[Charles Link]

It looks like the author is using cgs units, where $E=Q/r^2$. Meanwhile "on it", refers to the charge $Q$ that is "on it", $Q=(4 \pi r_o^2) \sigma$. $\\$ Editing... Additionally $E=4 \pi \sigma$ at $r=r_o$ (i.e. $r$ just slightly greater than $r_o$.)

 

[Buffu]

It looks like the author is using cgs units, where $E=Q/r^2$. Meaning "on it", refers to the charge $Q=(4 \pi r_o^2) \sigma$. Editing... $E=4 \pi \sigma$ at $r=r_o$ (i.e. $r$ just slightly greater than $r_o$.)

Yes you are correct. It my mistake, he meant charge on it is $4\pi r^2_0\sigma$ not the electric field.

Thanks and sorry for this useless question :((.