Homework Help: Can't Find Error in Application of Gauss's Law

1. Jan 15, 2010

SaintsTheMeta

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

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Gauss's Law $$\oint\vec{E}\text{dA=}\frac{Q_{encl}}{\epsilon_{0}}$$

3. 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

Last edited: Jan 15, 2010
2. Jan 15, 2010

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?

Last edited: Jan 15, 2010
3. Jan 15, 2010

SaintsTheMeta

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...

4. Jan 15, 2010

vela

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

5. Jan 15, 2010

SaintsTheMeta

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!!