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Electrodynamics Potential from charged sphere. I am lost :/

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~Electrodynamics~ Potential from charged sphere. I am lost :/

Homework Statement



A sphere of radius R, centered at the origin, carries a charge density ρ(r,θ)=κ/r^2(R-2r)sin^2(θ).
κ is constant. Find exact potential.

Homework Equations



1/4∏ε∫ρ∂t/r

The Attempt at a Solution


Question and attempt via attachments :)
I do not think my solution is correct please help!
Thanks all for looking!
 

Attachments

Answers and Replies

  • #2
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Shouldn't it be ##\frac{...}{\sqrt{z^2+r^2-2zr\cos\theta}}## instead of ##\frac{...}{\sqrt{R^2+r^2-2Rr\cos\theta}}## ?

as ##\sqrt{(z-r\cos\theta)^2+r^2\sin^2\theta}=\sqrt{z^2+r^2-2rz\cos\theta}## is a distance from the point in the sphere to the point in z-axis
 
  • #3
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Okay, yeah that makes sense. So the numerator looked good? If possible could you please work it out? I MUST have the integration done correctly! Thank you for responding!
 
  • #4
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Yes, the numerator is correct, you get:

##V(z)=\frac{K}{4\pi\varepsilon}\int_0^{2\pi}\int_0^R\int_0^\pi\frac{(R-2r)\sin^3\theta}{\sqrt{z^2+r^2-2zr\cos\theta}}\,d\theta drd\varphi##

so try a substitution ##u=z^2+r^2-2zr\cos\theta##
 
  • #5
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Using maple I got a ridiculous answer (attached)

What am I doing wrong?
 

Attachments

  • #6
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I don't know what is wrong. I calculated it in such a way:

##V(z)=\frac{K}{4\pi\varepsilon}\int_0^{2\pi}\int_0^R\int_0^\pi\frac{(R-2r)\sin^3\theta}{\sqrt{z^2+r^2-2zr\cos\theta}}\,d\theta drd\varphi=\frac{K}{2\varepsilon}\int_0^R\int_0^\pi\frac{(R-2r)(1-\cos^2\theta)\sin\theta}{\sqrt{z^2+r^2-2zr\cos\theta}}\,d\theta dr=(*)\\
z^2+r^2-2zr\cos\theta=u\\
\cos\theta=\frac{r^2+z^2-u}{2zr}\\
\sin\theta\,d\theta=\frac{du}{2zr}\\
(*)=\frac{K}{2\varepsilon}\int_0^R\int_{(z-r)^2}^{(z+r)^2}\frac{(R-2r)\left(1-\left(\frac{r^2+z^2-u}{2zr}\right)^2\right)}{2zr\sqrt{u}}\,dudr=...##
 
  • #7
vela
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Okay, yeah that makes sense. So the numerator looked good? If possible could you please work it out? I MUST have the integration done correctly! Thank you for responding!
That's not how it works here. We don't do the problem for you. That's your job.

Using maple I got a ridiculous answer (attached)

What am I doing wrong?
Using Maple for one thing. The point of this problem really is to learn how to do the math.

You should recognize the expression
$$\frac{1}{\sqrt{z^2+r^2-2zr\cos\theta}}.$$ Expand it in terms of Legendre polynomials.
 
  • #8
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If I had a nickle every time someone says "I plugged it into maple". . .

Also, a table of integrals comes in very handy for EM problems.
 
  • #9
vanhees71
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One should also say that the correct formula for the potential everywhere is (in SI units)
[tex]V(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{4 \pi \epsilon_0 |\vec{x}-\vec{x}'|}.[/tex]
Here, however, I'd rather use the multipole expansion of the potential and solve Poisson's Equation
[tex]\Delta V=-\frac{1}{\epsilon} \rho.[/tex]
 

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