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SaintsTheMeta
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Can't find the error I made... Rudimentary problem i know but i can't find where my mistake is...
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
Gauss's Law [tex]\oint\vec{E}\text{dA=}\frac{Q_{encl}}{\epsilon_{0}}[/tex]
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]\tex{E(4}\pi\tex{r^{2})=}\frac{1}{\epsilon_{0}}\tex{(}\frac{r^{3}}{r_{0}^{3}}\tex{)Q}[/tex]
[tex]\tex{E=}\frac{1}{\tex{4}\pi\epsilon_{0}}\tex{(}\frac{r}{r_{0}^{3}}\tex{)Q}[/tex]
[tex]\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[/tex]
[tex]{E=}\frac{8}{3\epsilon_{0}}{r}[/tex]
[tex]{E=3.0}\cdot{10^{11}r}\hspace{1 pc}\frac{N\cdot\tex{m}}{C}[/tex]
so far so good i think
now for the internal part...
[tex]{E(4}\pi\tex{r^{2})}=}\frac{Q}{\epsilon_{0}}[/tex]
[tex]{E=}\frac{1}{4\pi\tex{r^{2}}\epsilon_{0}}{(}\frac{4}{3}\pi\tex{0.06^{3})(-5)[/tex]
[tex]{E=}\frac{-0.00108}{3\cdot\tex{(8.9\cdot{10^{-12})r^{2}}}}[/tex]
[tex]{E=-4.1}\cdot{10^{7}}\frac{1}{r^{2}}[/tex]
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
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 [tex]\oint\vec{E}\text{dA=}\frac{Q_{encl}}{\epsilon_{0}}[/tex]
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]\tex{E(4}\pi\tex{r^{2})=}\frac{1}{\epsilon_{0}}\tex{(}\frac{r^{3}}{r_{0}^{3}}\tex{)Q}[/tex]
[tex]\tex{E=}\frac{1}{\tex{4}\pi\epsilon_{0}}\tex{(}\frac{r}{r_{0}^{3}}\tex{)Q}[/tex]
[tex]\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[/tex]
[tex]{E=}\frac{8}{3\epsilon_{0}}{r}[/tex]
[tex]{E=3.0}\cdot{10^{11}r}\hspace{1 pc}\frac{N\cdot\tex{m}}{C}[/tex]
so far so good i think
now for the internal part...
[tex]{E(4}\pi\tex{r^{2})}=}\frac{Q}{\epsilon_{0}}[/tex]
[tex]{E=}\frac{1}{4\pi\tex{r^{2}}\epsilon_{0}}{(}\frac{4}{3}\pi\tex{0.06^{3})(-5)[/tex]
[tex]{E=}\frac{-0.00108}{3\cdot\tex{(8.9\cdot{10^{-12})r^{2}}}}[/tex]
[tex]{E=-4.1}\cdot{10^{7}}\frac{1}{r^{2}}[/tex]
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
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