Register to reply

The Electric Field of a Uniformly Charged Disk

by Skoth
Tags: charged, disk, electric, field, uniformly
Share this thread:
Skoth
#1
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!
Phys.Org News Partner Science news on Phys.org
Scientists develop 'electronic nose' for rapid detection of C. diff infection
Why plants in the office make us more productive
Tesla Motors dealing as states play factory poker
SammyS
#2
Mar25-12, 08:31 PM
Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,819
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]


Register to reply

Related Discussions
Electric Field of a uniformly charged rod Introductory Physics Homework 7
Uniformly-charged disk electric field at ANY point (i.e., off-axis)? Advanced Physics Homework 2
Electric field due to a uniformly charged rod Introductory Physics Homework 1
Electric field at tip of uniformly charged cone Advanced Physics Homework 4
Electric Charge on a uniformly charged disk Introductory Physics Homework 1