Find E, circular disk. Radial component of E question.

In summary, the conversation discusses a problem from Sadiku's Elements of Electromagnetism involving a circular disk with uniform charge and the demonstration of the electric field at a specific point. There is a question about why the integration of the differential charges along the radial direction is zero, and the answer is that it is due to the cancellation of electric field vectors from charges at different angles on the disk. The conversation also includes a clarification about the use of cylindrical coordinates and the need to convert to Cartesian coordinates for a more straightforward integration.
  • #1
jfierro
20
1
Hi, I'm not sure this falls under the homework category, but I posted this here because I think this is more of an insight question than a question for help on how to solve the excercise (which I have).

I am reading Sadiku's Elements of Electromagnetism. There is an exercise reads something like this:

A circular disk of radius a has a uniform charge [tex]\rho_s \frac{C}{m^2}[/tex]. If the disk is on the plane z = 0 with its axis along the z axis,

a) Demonstrate that at point (0, 0, h)

[tex]\textbf{E} = \frac{\rho_s}{2\epsilon_0}\left\{ 1 - \frac{h}{\sqrt{h^2 + a^2}} \right\} a_z [/tex]

This is a visual representation, if lame, of the problem:

http://img191.imageshack.us/img191/1258/chargeddisk.gif [Broken]

Well, using Coulomb's I get to:

[tex]\textbf{E} =
\frac{\rho_s}{ 4\pi\epsilon_0 } \int_0^a\int_0^{2\pi}
\frac{ -r a_r + h a_z }
{ \left( r^2 + h^2 \right)^\frac{3}{2} }rd\theta dr[/tex]

Now, I recall reading in the book that the individual contributions to the E field along the [tex]a_r[/tex] (radial) direction sum up to zero, so I just discarded that vectorial component in the integral above and got easily to the demonstration. However, my question is:

Why is it that the integration of the differential charges along the radial direction [tex]a_r[/tex] is 0? I tried integrating:

[tex]
\int_0^a
\frac{ -r \textbf{a_r} }
{ \left( r^2 + h^2 \right)^\frac{3}{2} }rdr
[/tex]

Lets say the disk had radius 1, and we want to know the magnitude of the radial component at point (0, 0, 1), so that the integral becomes:

[tex]
-\int_0^1
\frac{ r^2 \textbf{a_r} }
{ \left( r^2 + 1 \right)^\frac{3}{2} }dr
[/tex]

But that, according to mathematica, is:

[tex]
\frac{1}{\sqrt{2}} - ArcSinh[1]
[/tex]

(Command used: Integrate[-(x^2)/((x^2 + 1^2)^(3/2)), {x, 0, 1}])

Which is not 0 o_O

Thanks in advance.
 
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  • #2
Sorry for the double post, I created a new thread while fixing some tex formatting issues... If a moderator reads this please delete the other thread.
 
  • #3
There is a \phi dependence that you are missing. Ignore the math and just visualize the electric field vector. The field vector from the charge at \phi=0 and some arbitrary radius on the disc is going to point up along the z-axis and away from the source point at the observation point on the z-axis. Now look at the field vector from the charge at \phi=\pi at the same radius. It will have the same magnitude and angle with the z-axis, the only difference is that it will be pointing in the opposite \rho direction. The \rho component of the electric field from a charge at \phi will be canceled out by the charge at \phi+\pi. So you need to double check your integrals because it is easy to demonstrate this to yourself conceptually.
 
  • #4
Thanks for your response, I guess it is easy to visualize it conceptually, the same goes for an infinite plane at any point perpendicular to it. However, There is something inside of me asking for some math to back it up :). So, if we add the \phi dependence, the integral for the radial component becomes (I'll use \rho and \phi since it seems it's more standard than r and \theta ):

20\vec{a_\rho}%20}%20{%20\left(%20\rho^2%20+%20h^2%20\right)^\frac{3}{2}%20}\rho%20d\phi%20d\rho.gif


Integrating on \phi:

rac{%20-\rho^2%20\vec{a_\rho}%20}%20{%20\left(%20\rho^2%20+%20h^2%20\right)^\frac{3}{2}%20}d\rho.gif


Which will be zero only if:

^a%20\frac{%20-\rho^2%20}%20{%20\left(%20\rho^2%20+%20h^2%20\right)^\frac{3}{2}%20}d\rho%20=%200.gif


Could you help me find out what I am missing math-wise?
 
  • #5
The problem is that cylindrical coordinates is inadequate for the integral. What happens if you try to sum the vectors x-hat and -x-hat when represented in cylindrical coordinates? You do not get the correct answer by just doing a naive addition. Convert it to cartesian coordinates for a no-brainer integration. In this case:

[tex] \vec{E} = \frac{\rho_s}{ 4\pi\epsilon_0 } \int_0^a\int_0^{2\pi} \frac{ \rho\cos\phi \hat{x} + \rho\sin\phi\hat{y} } { \left( \rho^2 + h^2 \right)^\frac{3}{2} }\rho d\phi d\rho [/tex]

It is obvious now that the integration over \phi will be zero.
 
Last edited:
  • #6
It is indeed very clear now that the contributions along the rho axis sum up to zero. Thanks :).
 

1. What is the formula for finding the radial component of E for a circular disk?

The formula for finding the radial component of E for a circular disk is E_radial = (1/4πε)*(σR/√(R^2+z^2)) where ε is the permittivity of free space, σ is the surface charge density, R is the radius of the disk, and z is the distance from the center of the disk to the point where the radial component is being measured.

2. How do you determine the direction of the radial component of E for a circular disk?

The direction of the radial component of E for a circular disk is always perpendicular to the surface of the disk, pointing away from the center of the disk. This can be determined by using the right-hand rule, where the fingers of the right hand point in the direction of the electric field lines and the thumb points in the direction of the radial component.

3. What is the significance of the radial component of E for a circular disk?

The radial component of E for a circular disk represents the strength and direction of the electric field at any point on the disk's surface. It is an important quantity in understanding the behavior of electric fields in 3-dimensional systems and is used in various applications such as calculating the force experienced by a charged particle on the disk's surface.

4. How does the radial component of E change as the distance from the center of the disk increases?

The radial component of E decreases as the distance from the center of the disk increases. This is because the surface area of the disk increases with distance, resulting in a decrease in the surface charge density. Additionally, the inverse-square law states that the electric field strength decreases with the square of the distance from the source.

5. Can the radial component of E for a circular disk be negative?

Yes, the radial component of E for a circular disk can be negative. This occurs when the direction of the electric field is pointing towards the center of the disk, which can happen if the disk has a negative surface charge density. It is important to note that the magnitude of the radial component is always positive, but the direction can be either towards or away from the center of the disk.

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