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## Homework Statement

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.

## Homework 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]

## 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!