The Electric Field of a Uniformly Charged Disk

In summary, we are calculating the electric field at a point P located along the central perpendicular axis of a disk with radius R and uniform surface charge density σ. Using the equations for electric field due to a continuous charge distribution and surface charge density, we set up the formula Ex = kex\piσ\int\frac{2rdr}{(r^{2}+x^{2})^{3/2}} and then rewrite it as kex\piσ\int(r^{2}+x^{2})^{-3/2}d(r^{2}). The d(r2) refers to an infinitesimal value of r2 and this is a normal operation in calculus. We can treat any integration by way of the power
  • #1
Skoth
7
0

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.
electricfieldofauniformlychargeddisk.jpg


Homework 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]

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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  • #2
Skoth said:

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.
electricfieldofauniformlychargeddisk.jpg


Homework 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]

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]
 

1. How is the electric field of a uniformly charged disk calculated?

The electric field of a uniformly charged disk can be calculated using the equation E = σ/2ε0 where σ is the surface charge density and ε0 is the permittivity of free space.

2. Is the electric field inside and outside of a uniformly charged disk the same?

No, the electric field inside and outside of a uniformly charged disk is not the same. The electric field inside the disk is zero, while outside the disk it follows an inverse square law and decreases as the distance from the disk increases.

3. Does the size of the disk affect the strength of the electric field?

Yes, the size of the disk does affect the strength of the electric field. The larger the surface area of the disk, the stronger the electric field will be.

4. What happens to the electric field if the charge on the disk is doubled?

If the charge on the disk is doubled, the electric field will also double. This is because the electric field is directly proportional to the charge.

5. Can the electric field of a uniformly charged disk be negative?

Yes, the electric field of a uniformly charged disk can be negative. This can happen if the charge on the disk is negative or if the electric field is directed towards the disk instead of away from it.

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