# 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!!