Electric field of a uniformly polarized sphere

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SUMMARY

The discussion focuses on calculating the electric field of a uniformly polarized sphere with radius R. The relevant equation for potential is given as V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \oint_{S} \frac{\sigma_{b}}{r} da' + \int_{V} \frac{\rho_{b}}{r} d\tau', where the surface charge density is defined as \sigma_{b} = P \cos\theta. The user correctly identifies the need to integrate over the angles, but misuses the variable \theta, which represents the angle between r and R. Proper notation and limits for integration are crucial for accurate calculations.

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stunner5000pt
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Homework Statement


Find the electric field of a uniformly polarized sphere of radius R

Homework Equations


[tex]V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \oint_{S} \frac{\sigma_{b}}{r} da' + \int_{V} \frac{\rho_{b}}{r} d\tau'[/tex]

The Attempt at a Solution


well obviously there is no volume charge density rho
but there is a surface charge density
[tex]\sigma_{b} = P \cos\theta[/tex]

now to calculate the potentail we got to use that above formula
Suppose r > R

then
[tex]V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \int \frac{P \cos\theta}{r} da'[/tex]

now the squigly r is found using the cosine law right...?

[tex]r = \sqrt{R^2 + r^2 - 2Rr\cos\theta}[/tex]
and
[tex]da' = R^2 \sin\theta d\theta d\phi[/tex]
is that right?
and the limits of integrate for the theta would be from 0 to pi
and for the phi is 0 to 2pi??

thanks for your help
(o by the way how do i put the squigly r??)
 
Last edited:
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Squigly r ??
Did you mean [tex]\tilde{r}[/tex] ?

Your solution is basically correct, but you have abuse the usage of [tex]\theta[/tex]. Notice the [tex]\theta[/tex] in [tex]\tilde{r} = \sqrt{R^2 + r^2 - 2Rr\cos\theta}[/tex] is represecting the angle between r and R. It is not the same [tex]\theta[/tex] in the rest of your equations... you should not treat it like a variable and integrate over it...
 
Last edited:
chanvincent said:
Squigly r ??
Did you mean [tex]\tilde{r}[/tex] ?

Your solution is basically correct, but you have abuse the usage of [tex]\theta[/tex]. Notice the [tex]\theta[/tex] in [tex]\tilde{r} = \sqrt{R^2 + r^2 - 2Rr\cos\theta}[/tex] is represecting the angle between r and R. It is not the same [tex]\theta[/tex] in the rest of your equations... you should not treat it like a variable and integrate over it...

sorry about the slopppy notation...

i shouldve put the primes
 

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