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The Electric Field of a Uniformly Charged Disk 
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#1
Mar2512, 02:00 PM

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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. 2. Relevant equations Electric field due to a continuous charge distribution: k_{e}[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: E_{x} = k_{e}x[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: k_{e}x[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(r^{2}) refers to an infinitesimal value of r^{2}. 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 ∫dx^{2}? More generally, could we treat any integration by way of the power rule like so: ∫x^{m}d(x^{n}), with the solution being [itex]\frac{x^{m+n}}{m+n}[/itex] + C? Or am I completely misinterpreting the calculation? Thanks! 


#2
Mar2512, 08:31 PM

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


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