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Electric Flux through a surface.

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider a spherical region V of radius R without any charge and a point charge q outside the spherical region at a distance d from the center as shown in the figure. Evaluate explicitly the net electric flux out of the surface S which bounds the spherical region V.

    2. Relevant equations
    Electric Field of a Point Charge (k=1):
    [tex]\vec{E}=\frac{q}{r^2}\hat{r}[/tex]

    Electric Flux out of a Surface:
    [tex]\Phi_{E}=\int_{S}\vec{E}\cdot d\vec{\sigma}[/tex]

    3. The attempt at a solution
    From Gauss' Law, I know the answer should be zero since there is no charge enclosed by the sphere. The main problem with solving this by using the surface integral instead, is that I can only place either the sphere or the charge at the origin, which yields a fairly complicated integral.

    If I consider the charge at the origin, then [tex]\vec{E}=\frac{q}{r^2}\hat{r}[/tex], and [tex]d\vec{\sigma}=R^2sin\theta d\Phi d\theta\hat{R}[/tex]

    If I place the sphere at the origin, then [tex]d\vec{\sigma}=R^2sin\theta d\Phi d\theta\hat{r}[/tex], but the electric field is then [tex]\vec{E}=\frac{q}{r^2}\hat{r'}[/tex]

    Here, due to the spherical symmetry, the integral can be simplified to [tex]\Phi_{E}=2\pi qR^2\int^{\pi}_{0}\frac{sin\theta}{r^2}(\hat{r}\cdot \hat{r'})d\theta[/tex]

    For some point on the circle (R, θ), r²=R²+d²-2Rdcosθ, so [tex]\Phi_{E}=2\pi qR^2\int^{\pi}_{0}\frac{sin\theta d\theta}{R^2²+d^2-2Rdcos\theta}(\hat{r}\cdot \hat{r'})[/tex]

    The problem for me is the dot product term, since both are unit vectors, it is simply the cosine of the angle between the to vectors. However, using the law of cosines again gives
    [tex]cos\theta '=\frac{R^2+r^2-d^2}{2Rr}=\frac{R-dcos\theta}{\sqrt{R^2²+d^2-2Rdcos\theta}}[/tex]

    Then the integral is [tex]\Phi_{E}=2\pi qR^2\int^{\pi}_{0}\frac{sin\theta (R-dcos\theta )d\theta}{(R^2²+d^2-2Rdcos\theta)^{3/2}}[/tex]

    The first integral, [tex]\int^{\pi}_{o}\frac{Rsin\theta d\theta}{(R^2²+d^2-2Rdcos\theta)^{3/2}}=\frac{-1}{d\sqrt{R^2²+d^2-2Rdcos\theta}}\right|^{\pi}_{0}=\frac{-1}{d\sqrt{R^2²+d^2-2Rdcos\pi}}+\frac{1}{d\sqrt{R^2²+d^2-2Rdcos0}}[/tex]
    [tex]=\frac{-1}{d\sqrt{R^2²+d^2+2Rd}}+\frac{1}{d\sqrt{R^2²+d^2-2Rd}}=\frac{-1}{d\sqrt{(R-d)^2}}+\frac{1}{d\sqrt{(R+d)^2}}=\frac{1}{d(R+d)}+\frac{-1}{d(R-d)}=\frac{-2}{(R^2-d^2)}[/tex]

    I assume the above integral is correct.

    The problem comes from the second integral, with the sinθcosθdθ in the numerator, it is not of a particular form I am familiar with, and there seem to be no obvious substitutions in order to integrate it. Is this solution on the right track, or is there something I missed that may have been easier? Thanks.
     
  2. jcsd
  3. Sep 20, 2009 #2

    gabbagabbahey

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    So far, so good!:approve:

    Not quite, if the point charge is outside the sphere, then [itex]d>R[/itex] and hence

    [tex]\sqrt{(R-d)^2}=d-R\neq R-d[/tex]

    Just make the substitution [itex]x=\cos\theta[/itex] and then integrate by parts.
     
  4. Sep 20, 2009 #3
    Adjusting for the square root mistake, then the first result will be
    [tex]\frac{2R}{d(R^2-d^2)}[/tex]

    For the secont integral I get
    [tex]\frac{\sqrt{R^2+d^2-2Rdcos\theta}}{2R^2d^2}+\frac{R^2+d^2}{2R^2d^2\sqrt{R^2+d^2-2Rdcos\theta}}\right|^{\pi}_{0}=\frac{R^2+d^2-Rdcos\theta}{R^2d^2\sqrt{R^2+d^2-2Rdcos\theta}}\right|^{\pi}_{0}[/tex]
    [tex]=\frac{R^2+d^2+Rd}{2R^2d^2\sqrt{R^2+d^2+2Rd}}-\frac{R^2+d^2-Rd}{R^2d^2\sqrt{R^2+d^2-2R}}=\frac{R^2+d^2+Rd}{2R^2d^2\sqrt{(R+d)^2}}-\frac{R^2+d^2-Rd}R^2d^2\sqrt{(R-d)^2}}=\frac{R^2+d^2+Rd}{R^2d^2(R+d)}-\frac{R^2+d^2-Rd}{R^2d^2(d-R)}[/tex]
    [tex]=\frac{R^2((d-R)-(d+R))+d^2((d-R)-(d+R))+Rd((d-R)+(d+R))}{R^2d^2(d+R)(d-R)}=\frac{R^2(-2R)+d^2(-2R)+Rd(2d)}{R^2d^2(d^2-R^2)}[/tex]
    [tex]=\frac{-2R^3}{R^2d^2(d^2-R^2)}=\frac{-2R}{d^2(d^2-R^2)}=\frac{2R}{d^2(R^2-d^2)}[/tex]

    This term is then multiplied by -d and added to the previous getting zero. This is what I expected from Gauss Law. I really can appreciate Gauss' Law in this case, it greatly simplifies this type problem.
     
  5. Sep 20, 2009 #4

    gabbagabbahey

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    Indeed!:smile:
     
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