Solve Tricky Double Integral for Charge Density Distribution in Symmetric Disc

In summary, the conversation discusses the attempt to model the potential field in and around a symmetrically charged disc with an exponentially decreasing charge density. The proposed method involves solving a double integral and using substitution of variables, but the result has not been successful and a symbolic solution is preferred. There is also a discussion about the accuracy of the proposed integral.
  • #1
TheMan112
43
1

Homework Statement


I'm trying to model the potential field in and around a symmetrically charged disc where the charge density drops exponentially from the center.

Homework Equations


This can be done by solving the double integral:

[tex]\int ^{2 \pi} _{0} \int ^{\infty} _{0} \frac{r e^{-r/b} (a-r sin \theta)}{(r^2 + a^2 - 2 r a cos \theta)^{3/2}} dr d \theta[/tex]

a is the observation radius, b is the scalar length of the charge density distribution, r and theta are polar coordinates. So essentially I'm looking for a result of the form V(a).

The Attempt at a Solution


I've been trying to attack this using substitution of variables without much result. It could be done numerically but I would greatly prefer a symbolic solution.
 
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  • #2
TheMan112 said:
I'm trying to model the potential field in and around a symmetrically charged disc where the charge density drops exponentially from the center.

Huh?:confused: Are you modeling the electrostatic potential, or the electric field? Does your disk extend out to infinity? Are you only looking the model the potential (or field) at point on the disk, or also at points not on the disk?



[tex]\int ^{2 \pi} _{0} \int ^{\infty} _{0} \frac{r e^{-r/b} (a-r sin \theta)}{(r^2 + a^2 - 2 r a cos \theta)^{3/2}} dr d \theta[/tex]

Even if you are just looking for the electric field at points on the disk, this integral doesn't look quite right to me... how did you get [itex]a-r\sin\theta[/itex] in your numerator?
 

Related to Solve Tricky Double Integral for Charge Density Distribution in Symmetric Disc

What is a double integral?

A double integral is a type of mathematical operation that calculates the volume under a surface given by a function of two variables. It is represented by two integral signs and is used to find the area between two curves or the volume of a solid with curved sides.

What makes a double integral tricky?

A double integral can be tricky because it involves integrating over two variables, which can lead to complicated calculations. Additionally, the limits of integration can be difficult to determine and may require using different approaches, such as changing the order of integration.

What are some techniques for solving tricky double integrals?

Some common techniques for solving tricky double integrals include using symmetry, changing the order of integration, and using substitution to simplify the integrand. It may also be helpful to sketch the region of integration and use geometric reasoning to determine the limits.

How is a double integral used in science?

Double integrals are used in a variety of scientific fields, including physics, engineering, and economics. They are particularly useful for calculating areas, volumes, and averages, and can be applied to model real-world problems.

What are some real-world applications of double integrals?

Double integrals are used in many real-world applications, such as calculating the center of mass of an object, determining the force exerted on a surface by a fluid, and finding the expected value of a random variable. They are also used in fields like signal processing and image processing to analyze and manipulate data.

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