The Electric Field of a Uniformly Charged Disk


by Skoth
Tags: charged, disk, electric, field, uniformly
Skoth
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Mar25-12, 02:00 PM
P: 7
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:

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!
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SammyS
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Mar25-12, 08:31 PM
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Quote Quote by Skoth View Post
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:

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!
[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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