Electric Potential: Griffith's EM Problem 3.26 Solution

In summary, the conversation discusses finding the approximate potential for points on the z axis, far from a sphere with a given charge density. The approach considered is using the multipole expansion formula, and after some integration, the integral with respect to theta is found to be -pi/8 for n=2, and -pi/64 for n=4. The sum itself is considered, with only the n=2 term being significant due to the others being too small. The final answer includes both dipole and quadropole terms, but the quadropole term is much smaller than the dipole term at large z and can be neglected for an approximation.
  • #1
stunner5000pt
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Griffith's EM Problem 3.26
A sphere of radius R centered at the origin carries a cahrge density
[tex] \rho(r,\theta) = k \frac{R}{r^2} (R - 2r) \sin \theta [/tex]

where k is constant and r and theta are teh usual spherical coordinates. Find teh approximate potentail for the points on the z axis, far from the sphere

i know ican do this with Laplace's equation but i wnana do it with the multipole expansion formula

[tex] V(\vec r) = \frac{1}{2 \pi \epsilon_{0}} \sum_{n=0}^{\infty} \frac{1}{r^{n+1}} \int (r')^n P_{n} (\cos \theta') \rho (r') d\tau' [/tex]

so about taht integral

[tex] kR \int_{0}^{R} (r')^n \frac{R-2r'}{r'^2} r'^2 dr' \int_{0}^{2\pi} P_{n} (\cos \theta) \sin^2 \theta d\theta \int_{0}^{\pi} \phi [/tex]

after some integration i found that the integral with respect to theta for
n =2 is -pi/8 dipole term
n = 4 = -pi/64 quadropole term

for n = 1, 3, and 5 is zero
so we end up with
[tex] \pi kR \frac{1 \pi}{8} \int_{0}^{R} (r')^n \frac{R-2r'}{r^2} r^2 dr [/tex]

[tex] \pi kR \frac{1 \pi}{8} \left[ \frac{-R(R)^{n+1}}{n+1} + \frac{2 R^{n+2}}{n+2} \right] [/tex]

now looking at hte sum intself

we consider only n = 2 since others are too small
[tex]\frac{-1 \pi^2}{8} \frac{KR^{n+3}}{r^{n+1}} \left( \frac{1}{n+1} - \frac{2}{n+2} \right) [/tex]

[tex] \frac{kR^5}{r^3} \pi^2 \frac{1}{48} [/tex]

is this fine?

should the answer include both dipole and quadropole terms?? Since this a sphere it makes sense to have that.
thank you for your help!
 
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  • #2
It did ask for an approximation, and the quadrapole term is way smaller than the dipole at large z, so I would leave it.

(I did not check your calculations. I assume you did it OK)
 
  • #3
Your integration limits for [tex]\phi[/tex] and [tex]\theta[/tex] should be switched.
 

FAQ: Electric Potential: Griffith's EM Problem 3.26 Solution

1. What is electric potential?

Electric potential is a measure of the potential energy per unit charge at a certain point in space. It is typically measured in volts (V) and is a fundamental concept in the study of electricity and magnetism.

2. What is Griffith's EM Problem 3.26?

Griffith's EM Problem 3.26 is a well-known problem in the field of electromagnetism that deals with finding the electric potential inside and outside of a charged spherical shell.

3. What is the solution to Griffith's EM Problem 3.26?

The solution to Griffith's EM Problem 3.26 involves using the equations for electric potential due to a point charge and a charged spherical shell to find the potential at different points in space.

4. Why is Griffith's EM Problem 3.26 important?

Griffith's EM Problem 3.26 is important because it is a commonly used example in textbooks and courses to illustrate the concept of electric potential and how it is affected by different types of charge distributions.

5. How is the solution to Griffith's EM Problem 3.26 used in practical applications?

The solution to Griffith's EM Problem 3.26 can be used in practical applications such as designing capacitors and other electronic devices, understanding the behavior of electric fields in conductors, and predicting the behavior of charged particles in electric fields.

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