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Electrodynamics Potential from charged sphere. I am lost :/

  1. Sep 25, 2013 #1
    ~Electrodynamics~ Potential from charged sphere. I am lost :/

    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    1/4∏ε∫ρ∂t/r

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

    Attached Files:

  2. jcsd
  3. Sep 25, 2013 #2
    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
     
  4. Sep 25, 2013 #3
    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!
     
  5. Sep 25, 2013 #4
    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##
     
  6. Sep 25, 2013 #5
    Using maple I got a ridiculous answer (attached)

    What am I doing wrong?
     

    Attached Files:

  7. Sep 26, 2013 #6
    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=...##
     
  8. Sep 26, 2013 #7

    vela

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    That's not how it works here. We don't do the problem for you. That's your job.

    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.
     
  9. Sep 26, 2013 #8
    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.
     
  10. Sep 27, 2013 #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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