Fourier Tranform of electric dipole charge density

In summary, the conversation discusses a problem involving two charges, one positive and one negative, at specific coordinates. The task is to use Fourier transform to find the charge density. The equations and attempt at a solution are also mentioned, including the use of the delta function and the potential of the charges. The conversation ends with a discussion about the usefulness of the transform of the charge density.
  • #1
naftali
31
0

Homework Statement


Hi,

This is supposed to be simple, so I guess I miss something..
We have charge q at x1=d*cos (w*t), y=0, z=0. and charge -q at
x2=-d*cos (w*t), y=0, z=0. I need to do Fourier transform to the charge density.

Homework Equations


The Fourier transform is : [itex]\rho_{\omega} = \int \rho (r,t)e^{i\omega t} dt [/itex]


The Attempt at a Solution


I first try only for the first charge. We have [itex]\rho (r,t) = \delta(x-d\cdot cos (\omega t))\cdot\delta(y)\cdot\delta(z) [/itex]
But I don't know how to do the intergral :
[itex]\rho_{\omega} = \int \delta(x-d\cdot cos (\omega t))\cdot\delta(y)\cdot\delta(z)e^{i\omega t} dt [/itex]
Any ideas? Do I really need to do this integral?
 
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  • #2
Generally, if you have an integral over a delta function of another function, you can use the relation

[tex]\delta(f(t)) = \sum_{j}\frac{\delta(t-t^\ast_j)}{|f'(t^\ast_j)|},[/tex]
where the [itex]t^\ast_j[/itex] are the solutions of [itex]f(t^\ast) = 0 [/itex]. Note that this relation doesn't work if the derivative is zero at a solution.
 
  • #3
Hi,

Thanks for your answer.
Actually the question is to find the potential of this charges far from the origin,so I guess I should transform the dipole moment vector only..
Using the identity for the delta function I get weird things like exp(arccos (x/d))...
It is just interesting if the transform of the charge density could be useful.
 

Related to Fourier Tranform of electric dipole charge density

1. What is the Fourier Transform of electric dipole charge density?

The Fourier Transform of electric dipole charge density is a mathematical tool used to convert a function of time or space into a function of frequency or wavenumber. In the case of an electric dipole charge density, the Fourier Transform can be used to analyze the distribution of electric charges along an axis.

2. Why is the Fourier Transform of electric dipole charge density important?

The Fourier Transform of electric dipole charge density is important because it allows us to understand the behavior and properties of electric dipoles in terms of their frequency components. This can be useful in a variety of applications, such as signal processing and electromagnetic wave analysis.

3. How is the Fourier Transform of electric dipole charge density calculated?

The Fourier Transform of electric dipole charge density can be calculated using the formula: F(k) = ∫e^(ikx)ρ(x)dx, where F(k) is the transformed function, k is the wavenumber, and ρ(x) is the original function representing the charge density along the x-axis.

4. What are some properties of the Fourier Transform of electric dipole charge density?

Some properties of the Fourier Transform of electric dipole charge density include linearity, time and frequency shifting, and convolution. These properties can be used to simplify calculations and better understand the behavior of electric dipoles.

5. How does the Fourier Transform of electric dipole charge density relate to the electric field?

The Fourier Transform of electric dipole charge density can be used to calculate the electric field generated by an electric dipole. By taking the inverse Fourier Transform of the transformed function, we can obtain the electric field as a function of position or time.

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