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Volume Charge Density of an infinite cylinder.

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  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data


    An infinite cylinder of radius [tex]\textbf{R}[/tex] has a linear charge density [tex]\lambda[/tex]. The volume charge density [tex]\frac{C}{m^{3}}[/tex] within the cylinder [tex](r\leq R)[/tex] is [tex]\rho(r)=\frac{r\rho_{0}}{R}[/tex] where [tex]\rho_{0}[/tex] is a contant to be determined.

    The charge within a small volume [tex]d\textbf{V}[/tex] is [tex]dq=\rho d\textbf{V}[/tex]. The integral of [tex]\rho d \textbf{V}[/tex] over a cylinder of length [tex]\textbf{L}[/tex] is the total charge [tex]Q=\lambda L[/tex] within the cylinder. Use this fact to show that [tex]\rho_{0}=\frac{3\lambda}{2\pi R^{2}}[/tex].


    2. Relevant equations

    The volume of a cylinder is given as such:

    [tex]V=\pi r^{2}h[/tex]

    Where r is the radius of the cylinder, and h is the height of the cylinder.

    I suspect that the rest of the relevant equations have been included in the statement of the question.

    3. The attempt at a solution

    My strategy is this: find an integral with respect to r and incorporating [tex]\rho_{0}[/tex], and having performed the integration, solve for [tex]\rho_{0}[/tex].

    One might think of using an integral when one sees that the adding of all the little bits of charge equals Q, the total charge. The little bits of charge might be rewritten as such [tex]dq=\rho dV[/tex]. Therefore, the total integral [tex]\int{\rho dV}[/tex] must equal Q, the total charge. As V is volume, we should be able to rewrite this with respect to a coordinate such as r, which is the radius.

    I attempted to integrate the cylinder by regarding it as the sum of many thin "shells", of the thickness [tex]dr[/tex]. See the diagram below.
    View attachment forum.bmp
    Thereby, each shell will have height L, radius r, and thickness dr. Please note that R is the radius of the cylinder in its entirety, while r is simply the distance away from the centre of the cylinder to the beginning of the shell in question. Put differently, it's the radius of the "empty" cylinder inside the shell in question.

    I have difficulty in determining the volume of such a shell in such a way that I can use it in an integral. I keep getting powers on my differentiable bits, dr.

    Allow me to show:
    What would the volume be of such a shell?
    [tex]\begin{equation}
    \begin{align}
    dV&=(L\pi(r+dr^{2}))-(L\pi r^{2})=L\pi((r+dr)^{2}-r^{2}) \\
    &=L\pi(2r dr+dr^2)
    \end{align}
    \end{equation}[/tex]

    We are left with an undesirable integration:
    [tex]\begin{equation}
    \begin{align}
    L\pi\int{\rho(r)(2r dr+dr^{2})}\\
    \frac{\rho_{0}L\pi}{R}\int{r(2r dr +dr^{2})}
    \end{align}
    \end{equation}[/tex]

    It seems like I'm taking this from the wrong way. How can I solve this question using some sort of integration?

    Thank you very much for your time and consideration. Any guidance (and attending logic) towards a working integral would be great! The algebra leading to the final solution shouldn't be needed. Thanks again.
     
  2. jcsd
  3. Feb 3, 2010 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member



    Use the average of r(inside) and r(outside) as "r". If r(out)=r(in)+dr, your volume element is 2r *pi *h*dr .

    But anyway: when using a volume element, the higher order of the differentials are to be neglected.


    ehild
     
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