- #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
Thanks in advance.
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
Thanks in advance.
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