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The Electric Field of a Uniformly Charged Disk

  1. Mar 25, 2012 #1
    1. The problem statement, all variables and given/known data

    A disk of radius R has a uniform surface charge density σ. Calculate the electric field at a point P that lies along the central perpendicular axis of the disk and a distance x from the center of the disk.
    [​IMG]

    2. Relevant equations

    Electric field due to a continuous charge distribution:

    ke[itex]\int[/itex][itex]\frac{dq}{r^{2}}[/itex][itex]\hat{r}[/itex]

    Surface charge density:

    σ = [itex]\frac{Q}{A}[/itex]

    3. The attempt at a solution

    This is an example problem from my book, so I already know the solution. The real question I have is how they go about solving it. For the most part, it makes sense, but there is one small step in their calculation that I am confused by.

    They setup their formula as thus: Ex = kex[itex]\pi[/itex]σ[itex]\int[/itex][itex]\frac{2rdr}{(r^{2}+x^{2})^{3/2}}[/itex]. This I understand, but their next step is what throws me off. They rewrite this as:

    kex[itex]\pi[/itex]σ[itex]\int[/itex](r[itex]^{2}[/itex]+x[itex]^{2}[/itex])[itex]^{-3/2}[/itex]d(r[itex]^{2}[/itex])

    I assume that the d(r2) refers to an infinitesimal value of r2. Yet, what I'm confused by is if this is a normal operation in calculus (for I've never seen it before)? In other words, for any other integrations where you have, say, ∫xdx, could this be rewritten as ∫dx2? More generally, could we treat any integration by way of the power rule like so: ∫xmd(xn), with the solution being [itex]\frac{x^{m+n}}{m+n}[/itex] + C? Or am I completely misinterpreting the calculation?

    Thanks!
     
  2. jcsd
  3. Mar 25, 2012 #2

    SammyS

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    [itex]\displaystyle d\,\left(f(x)\right)=f'(x)\,dx[/itex]

    So, [itex]\displaystyle d(r^2)=\left(\frac{d}{dr}(r^2)\right)dr=2r\,dr\,.[/itex]
     
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