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Need someone to tell me if I got this done correctly

(a) Draw Gaussuian cylinder inside the black cylinder to find charge enclosed

$Q_{en} = Q(\frac{r}{a})^2$

Apply Gauss's Law

$E2\pi r \ell = \dfrac{Q\left ( \frac{r}{a} \right )^2}{\varepsilon_0}$

$E = \dfrac{Q\left ( \frac{r}{a} \right )^2}{2 \pi r \ell\varepsilon_0} = \dfrac{Qr}{2\pi \varepsilon_0 a^2\ell}$

$E = \dfrac{\rho r}{2 \varepsilon_0}$

(b) Basically same principle, but the total charge of the black cylinder

$E = \dfrac{\rho a^2}{2r \varepsilon_0}$

(c) 0 under static equilibrium

(d) The enclosed charge should be a sum of cylinders' charges.

$Q_{en} = \rho V + \lambda \ell = \rho \pi a^2 \ell + \lambda \ell = \ell ( \rho \pi a^2 + \lambda)$

Hence

$E2\pi r \ell = \dfrac{\ell ( \rho \pi a^2 + \lambda)}{\varepsilon_0} \implies E =\dfrac{ \rho \pi a^2 + \lambda}{2 \pi r \varepsilon_0} $

The term looks very awkward, so it's probably wrong.

Thank you for reading

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1 Answer 1

up vote 1 down vote accepted

As for a), note that $Q=\rho \pi a^2 l$.

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Wow I feel really stupid all of sudden. –  Hawk Aug 16 '12 at 18:13
    
Wait, if I sub in $Q=\rho \pi a^2 l$, I get answer for (b) –  Hawk Aug 17 '12 at 20:06
    
Looks like that –  akhmeteli Aug 17 '12 at 21:35
    
How can that happen? Doesn't make sense to me. cylinders aren't point charges. Did I do (d) right? –  Hawk Aug 18 '12 at 1:24
    
I don't understand what point charges have to do with that. And no, your (d) is not right. E.g., you should write $\rho \pi a^2$ instead of $\rho \pi r^2$. As for (b), again, maybe I misunderstood you, but the enclosed charge does not depend on the radius of the point in (b). By the way, I believe $R$ in your answers is the same as $a$. Another thing, $Q$ in your answers is not the enclosed charge in (a), but just a coefficient in the expression for the enclosed charge. –  akhmeteli Aug 18 '12 at 3:01

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