Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I think I am over thinking this question: A point charge of -2 microC is located in the center of a hollow sphere. The internal and external radius of the sphere is given by 6.5 and 9.5 cm, the charge density is 7.35 E(-4). Plot electric field vs. distance from the center of the sphere.

My thoughts were the region before r1 it would just be density*area, for r1 to r2 the value would be zero, and for a radius greater than r2 out of the sphere it would have an inverse 1/r^2 curve. Am I horribly off? Any help would be greatly appreciated, thanks

share|improve this question
3  
Hint: use Gauss's law. –  Marek Feb 2 '11 at 15:20
3  
This looks like a homework question –  gigacyan Feb 2 '11 at 15:20
    
gigacyan beat me to the retag :) –  Colin K Feb 2 '11 at 15:20
3  
What is there in between r1 and r2? some metal/conducting material? Where is that charge density measured? –  Georg Feb 2 '11 at 15:50
add comment

1 Answer

There is a charge at the center with charge, $q$, and a charged shell with finite width with inner and outer radii $a$ and $b$ respectively and charge density $\rho$, if I understand correctly.

Then there are three regions: $(0,a)\cup(a,b)\cup(b,\infty)$ combined that we have to find the field in and then plot. Using Gauss' law and focusing on the inner most region we have $$\oint E_i da_i = \frac{q_{\text{enc}}}{\epsilon_0}$$ Then because of the spherical symmetry we can write \begin{equation} E4\pi r^2=\frac{q}{\epsilon_0}\implies E_r=\frac{q}{4\pi\epsilon_0 r^2} \end{equation} Then for the middle region we can do the same thing but a little more explicit $$\oint E_i da_i =\frac{1}{\epsilon_0}\int \rho\, d\tau$$ For this region then we have $$E4\pi r'^2=\frac{\rho}{\epsilon_0}\int_{a}^{r'}d\tau =\frac{\rho}{\epsilon_0}\frac{4}{3}\pi (r'^3 - a^3)\implies E_r=\frac{\rho (r'^3 -a^3)}{3\epsilon_0 r'^2}$$ Then totally in that region we have $$E_r=\frac{\rho (r'^3 -a^3)}{3\epsilon_0 r'^2}+\frac{q}{4\pi\epsilon_0 r'^2}$$ Finally for completely outside the shell we simply note that the total charge is the charge inside, and the whole shell, which really means let $r'$ in the first term go to $b$, i.e. $$E_r=\frac{\rho (b^3 -a^3)}{3\epsilon_0 r'^2}+\frac{q}{4\pi\epsilon_0 r'^2}$$ since we don't get any more charge outside of the shell's radius.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.