Integrating a Difficult Integral: Electric Field of a Spherical Surface

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In summary, the electric field at a distance z from the center of a spherical surface that carries a uniform surface charge ##\sigma## is given by:\int_0^\pi \frac{(z-Rcos\theta)sin\theta d\theta}{(R^{2}+z^{2}-2Rzcos\theta)^{3/2}}
  • #1
ELB27
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Homework Statement


While trying to find the electric field at a distance z from the center of a spherical surface that carries a uniform surface charge ##\sigma## I got stuck with the following integral (which I'm quite sure is correct):
[tex]\int_0^\pi \frac{(z-Rcos\theta)sin\theta d\theta}{(R^{2}+z^{2}-2Rzcos\theta)^{3/2}}[/tex]

Homework Equations


The Attempt at a Solution


The only idea I had is to let ##u=cos\theta ; du=-sin\theta d\theta##,then the integral becomes:
[tex]\int_0^\pi \frac{(Ru-z)du}{(R^{2}+z^{2}-2Rzu)^{3/2}}[/tex]
and now I'm stuck. This looks very similar to a partial fractions problem except there is a square root in the denominator and if I understand correctly I'm not allowed to use partial fractions in this case.

Any help will be appreciated!
 
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  • #2
You want to integrate with respect to which variable?
 
  • #3
preferably ##\theta## but if my substitution of ##u## is of any good than ##u## will do also.
 
  • #4
That looks sort-of familiar ... if you're sure, I'll start from there.

Remember that z does not depend on u in your last integral, so it has form $$\int \frac{ax-b}{(c-dx)^{3/2}}\;\text dx$$
... your main problem is to get rid of the square-root in the denominator.
How would you normally do that?
 
Last edited:
  • #5
substitute [tex]x=\frac{c}{d}sinu[/tex]
EDIT: no, that's not correct, no 2nd power... perhaps multiply and divide by the square root. Am I allowed to use partial fractions with a square root in the numerator?
 
Last edited:
  • #6
You do want to get rid of the root before attempting partial fractions.
How about ##u^2=c-dx## or ##dx=c\sin^2\phi##

Aside: I probably should not have used "d" as a constant :(
 
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  • #7
Alright, substituting ##v^2=c-gx## (replacing ##d## with ##g##) I am finally able to integrate and simplify to get the desired result:
[tex]\int\frac{\frac{ac-av^{2}}{g}-b}{v^{3}}
\frac{-2v}{g}
dv = \int\frac{2a}{g^{2}}dv
- \int\frac{2ac}{g^{2}v^{2}}dv
+ \int\frac{2b}{gv^{2}}dv
=\frac{2av}{g^{2}}
+ \frac{2ac}{g^{2}v}
- \frac{2b}{gv}
= \frac{1}{z^{2}}\left(\frac{R+z}{|R+z|}-\frac{R-z}{|R-z|}\right)[/tex]

Thank you very much Simon!
 

1) What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to calculate the total amount of a quantity, such as distance or volume, when the rate of change of that quantity is known.

2) What is a problem with an integral?

A problem with an integral usually refers to a difficulty in solving a specific integral equation. This can occur when the equation is complex or when there is no known method to solve it.

3) How do you solve a problem with an integral?

Solving an integral equation involves applying various integration techniques, such as substitution or integration by parts, to simplify the equation and find its solution. If these techniques do not work, numerical methods can be used to approximate the integral.

4) What are some common mistakes when working with integrals?

Some common mistakes when working with integrals include forgetting to account for the limits of integration, incorrect use of integration rules, and forgetting to include the constant of integration when finding the antiderivative.

5) How are integrals used in real-world applications?

Integrals are used in a variety of real-world applications, such as calculating the area under a curve in physics or finding the volume of a solid object in engineering. They are also used in finance to calculate the total value of a continuous investment over time.

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