Can't Find Error in Application of Gauss's Law

[SaintsTheMeta]

Can't find the error I made... Rudimentary problem i know but i can't find where my mistake is...

Homework Statement

A nonconducting sphere is made of two layers. The innermost section has a radius of 6.0 cm and a uniform charge density of -5.0 C/m³. The outer layer has a uniform charge density of +8.0 C/m³ and extends from an inner radius of 6.0 cm to an outer radius of 12.0 cm. Determine the electric field for: ... (b) 6.0 cm < r < 12.0 cm

Homework Equations

Gauss's Law \oint\vec{E}\text{dA=}\frac{Q_{encl}}{\epsilon_{0}}

The Attempt at a Solution

Basically I would find the electric field due to the inner layer and then the outer layer and add them together.
Starting with the outer layer:
\tex{E(4}\pi\tex{r^{2})=}\frac{1}{\epsilon_{0}}\tex{(}\frac{r^{3}}{r_{0}^{3}}\tex{)Q}
\tex{E=}\frac{1}{\tex{4}\pi\epsilon_{0}}\tex{(}\frac{r}{r_{0}^{3}}\tex{)Q}
\tex{E=}\frac{1}{\tex{4}\pi\epsilon_{0}}\tex{(}\frac{1}{0.12^{3}}\tex{)(}\frac{4}{3}\pi\cdot\tex{0.12^{3})(8)r
{E=}\frac{8}{3\epsilon_{0}}{r}
{E=3.0}\cdot{10^{11}r}\hspace{1 pc}\frac{N\cdot\tex{m}}{C}
so far so good i think
now for the internal part...
{E(4}\pi\tex{r^{2})}=}\frac{Q}{\epsilon_{0}}
{E=}\frac{1}{4\pi\tex{r^{2}}\epsilon_{0}}{(}\frac{4}{3}\pi\tex{0.06^{3})(-5)
{E=}\frac{-0.00108}{3\cdot\tex{(8.9\cdot{10^{-12})r^{2}}}}
{E=-4.1}\cdot{10^{7}}\frac{1}{r^{2}}

however there is the problem... i don't see where my mistake was in that second part, but the book says the second part of the Electric Field from the core sphere should be -1.1x10^8/r^2...

so confused maybe its just too late (early) but any help is appreciated

 

[Gear300]

Does the book explicitly state that the electric field from the inner sphere is -1.1 x 10^8/r^2, or does it state the electric field between two radial distances?

 

[SaintsTheMeta]

Does the book explicitly state that the electric field from the inner sphere is -1.1 x 10^8/r^2, or does it state the electric field between two radial distances?

The book gives the answer as exactly:
-(1.1 x 10^8 N*m^2/C)/r^2 + (3.0 x 10^11 N/C*m)r

And as my outer shell matches the second part of their answer exactly, i can only assume that the first part is what my inner shell should amount to...

 

[vela]

Your calculation for the outer shell assumes you have a solid sphere, including the region r&lt;0.06~\mbox{m}. To compensate for that, the book used a charge density of -13 C/m^3 for the inner sphere.

 

[SaintsTheMeta]

Your calculation for the outer shell assumes you have a solid sphere, including the region r&lt;0.06~\mbox{m}. To compensate for that, the book used a charge density of -13 C/m^3 for the inner sphere.

OH! I see!

Yea for anyone else that might stumble here from google like i always do, I forgot to subtract out the inside of the outer sphere.

Thank you vela!