Potential outside a charged disc

In summary, the homework statement is that find the potential outside a thin, circular disc with charge q. The potential is given by U(r,z)=c(\lambda)e^{-\lambda z}J_0(\lambda r) where J_0 is the 0'th bessel function. The general solution then becomes usercontent{ \int_{-\infty}^{\infty}C(\lambda)J_0(\lambda r) d\lambda= \frac{q}{4\pi\epsilon r} } and if I work on the z-axis it becomes usercontent{ \int_{-\infty}^{\infty}C(\lambda)e^{-
  • #1
Azael
257
1

Homework Statement



Find the potential outside a thin, circular disc with charge q.

Homework Equations



Laplace's equation
DivgradU=0

The Attempt at a Solution



Since the problem implies symmetry around phi its obvious the solution doesn't depend on phi.

I separate the differential equation into U=A(r)B(z)

Then its easy to find the solutions
[tex]B(z)=c(\lambda)e^{-\lambda z}[/tex]
[tex]A(r)=J_0(\lambda r)[/tex]
where J_0 is the 0'th bessel function.

So I have

[tex]U(r,z)=c(\lambda)e^{-\lambda z}J_0(\lambda r)[/tex]

The general solution then becomes
[tex]\int_{-\infty}^{\infty}C(\lambda)e^{-\lambda z}J_0(\lambda r) d\lambda[/tex]

The problem is that I don't have the slighest clue on how to find C. I know that far away from the disc the potential should tend toward the potential outside a point charge.

So if we work along the r plane and put z=0 we have the equation, when r is large

[tex]\int_{-\infty}^{\infty}C(\lambda)J_0(\lambda r) d\lambda= \frac{q}{4\pi\epsilon r} [/tex]

If I instead work on the z axis I know that the potential is given by.

[tex]\int_{-\infty}^{\infty}C(\lambda)e^{-\lambda z} d\lambda= \frac{q}{2\pi\epsilon a^2}(\sqrt{z^2+a^2}-|z|) [/tex]

How do I transform this to solve c?:confused:
 
Last edited:
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  • #2
Azael;1270416 So if we work along the r plane and put z=0 we have the equation said:
\int_{-\infty}^{\infty}C(\lambda)J_0(\lambda r) d\lambda= \frac{q}{4\pi\epsilon r}[/tex]

If I instead work on the z axis I know that the potential is given by.

[tex]\int_{-\infty}^{\infty}C(\lambda)e^{\lambda z} d\lambda= \frac{q}{2\pi\epsilon a^2}k(\sqrt{z^2+a^2}-|z|) [/tex]

How do I transform this to solve c?:confused:

First, it should be something like

[tex] \int_{-\infty}^{\infty} C(\lambda) e^{-\lambda z} d\lambda= \frac{q}{2\pi\epsilon a^2}k(\sqrt{z^2+a^2}-|z|) [/tex]

Then, is "k" a Bessel function ? I'd say it looks that this second equation should be asking for a reverse Laplace transformation...

EDIT:Apparently the tex code is causing trouble. Just click the code to see the minor correction (a minus sign in the exp)
 
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  • #3
dextercioby said:
First, it should be something like

[tex] \int_{-\infty}^{\infty} C(\lambda) e^{-\lambda z} d\lambda= \frac{q}{2\pi\epsilon a^2}k(\sqrt{z^2+a^2}-|z|) [/tex]

Then, is "k" a Bessel function ? I'd say it looks that this second equation should be asking for a reverse Laplace transformation...

EDIT:Apparently the tex code is causing trouble. Just click the code to see the minor correction (a minus sign in the exp)

Ops, the k was just a typo and it shouldn't be there at all:blushing: Yoru right its a minus in the exp aswell. Edited my first post so its correct.

Is there any other way beside laplace transform to find C? Laplace transforms isn't even part of this class and hasnt been part of any previous class so its odd if we would be given a problem that require them.

In arfken there is a hint that the problem can be solved by Fourier transform but I haven't been able to come up with a way.:confused:
 
  • #4
I think the problem is easier in spherical coords.
Your result for the z-axis potential can be exanded to all angles using Legendre polynomials.
 
  • #5
Il have to give spherical coordinates a try tomorrow! Thanks for that tip didnt think about that.

But I hope I can figure it out in cylindrical coordinates aswell. I hate beeing stuck :(
 

Related to Potential outside a charged disc

What is potential outside a charged disc?

Potential outside a charged disc is the measure of the electric potential at a point outside a disc with a given charge. It is influenced by the electric field created by the disc's charge distribution.

How is potential outside a charged disc calculated?

Potential outside a charged disc is calculated using the equation V = kQ/r, where V is the potential, k is the Coulomb's constant, Q is the charge of the disc, and r is the distance from the center of the disc to the point of interest.

What is the effect of distance on potential outside a charged disc?

The potential outside a charged disc decreases as the distance from the disc increases. This is because the electric field strength decreases with distance, resulting in a weaker potential.

Can potential outside a charged disc be negative?

Yes, potential outside a charged disc can be negative. This occurs when the point of interest is located within the electric field of the disc, and the potential is influenced by both the disc's charge and the charges of other nearby objects.

How does the charge of the disc affect potential outside a charged disc?

The potential outside a charged disc increases as the charge of the disc increases. This is because a higher charge results in a stronger electric field, which in turn leads to a higher potential at a given distance from the disc.

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