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E field above charged spherical surface

  1. Mar 27, 2013 #1
    the calculation of the E field a distance z above a spherical surface of charge gives rise to this integral which can be done by partial fractions....I don't see this integral in Stewart's table
    did I not input correctly in Mathematica?

    Integrate [(z-R*u)/(R^2+z^2-2*z*R*u)^(3/2),{u,-1,1}]
    ConditionalExpression[(-R+z)/( z2)+/(z2 (R+z)),((Im[R] Re[R]+Im[z] Re[z])/(Im[z] Re[R]+Im[R] Re[z])1||Im[z] Re[R]+Im[R] Re[z]0||(Im[z] Re[R]+Im[R] Re[z]0&&((Im[R]+Im[z]) (Re[R]+Re[z])0||Im[R]3 Re[z]+Im[z] Re[R] (Im[z]2-Re[R]2+Re[z]2)Im[R] (Im[R] Im[z] Re[R]+Re[z] (Im[z]2-Re[R]2+Re[z]2))))||(Im[z] Re[R]+Im[R] Re[z]0&&((Im[R]+Im[z]) (Re[R]+Re[z])0||Im[R]3 Re[z]+Im[z] Re[R] (Im[z]2-Re[R]2+Re[z]2)Im[R] (Im[R] Im[z] Re[R]+Re[z] (Im[z]2-Re[R]2+Re[z]2)))))&&(R/z+z/RReals||Re[R/z+z/R]2||Re[R/z+z/R]-2)]
  2. jcsd
  3. Mar 28, 2013 #2


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    That looks really messy. You can use Gauss' law to get rid of the integral.
  4. Mar 28, 2013 #3


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    Does the surface have a uniform surface charge density σ? If so, you can choose the variable of integration so as not to have to use partial fractions. A suitable u-substitution makes it fairly easy to solve.

    (this sounds like Problem 2.7 in Griffiths :smile:)

    [added] OK, looking at my notes, I think I see where your integral came from. Hint: try a different definition of u, one that makes the denominator of the integrand really simple.
    Last edited: Mar 28, 2013
  5. Mar 28, 2013 #4
    when you do the spherical double integral from θ=0 to ∏, [itex]\phi[/itex]=0 to 2 ∏ are you summing the electrical forces from infinitesimal surface areas on the sphere?

    is this too simple for the general expression of the z component of the electrical field?


    where alpha is the angle between the z axis and the vector r

    I was assuming a uniform σ, but I'm not sure about this? if I do this double, 2nd integral substitution after the cosine substitution y=(Z^2+R^2+2zRx)^(1/2)
    the integral solution is-


    I think it can be done with partial fractions also. for the case that z>R, point is outside the sphere, z-R term=2 so that you have 4[itex]\pi[/itex]R[itex]^{2}[/itex]σ=Q, the surface charge is as if it is located at the center of the sphere in a single charge and for z<R the z-R term=0
    so it is a derivation of Gauss's law
    Last edited: Mar 28, 2013
  6. Mar 29, 2013 #5
    so when you calculate the electrical field from the sphere on a field point you get this beautiful result of the charge focused at the center of the sphere. And according to Gauss's Law the flux is the same regardless of the geometry of the closed surface and you can calculate the E field or flux density if the charge distribution is symmetric, but Coulumb's Law holds if the distribution is asymmetric? are there any other implications of this?
    Last edited: Mar 29, 2013
  7. Mar 30, 2013 #6


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    I'd say the most intuitive way is to integrate the electrostatic equations (written in Heaviside-Lorentz units and for the vacuum),
    [tex]\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=\rho.[/tex]
    The first equation tells us that the electric field has a scalar potential,
    [tex]\vec{E}=-\vec{\nabla} \cdot \Phi.[/tex]
    Gauß's Law then becomes the Poisson Equation,
    [tex]\Delta \Phi=-\rho.[/tex]
    In your case of a homogeneously charged spherical shell you have
    [tex]\rho(\vec{x})=\frac{Q}{4 \pi R^2} \delta(r-R),[/tex]
    where [itex]r=|\vec{x}|[/itex].

    The charge distribution is spherically symmetric and thus one can expect that the potential is a function of [itex]r[/itex] only. Writing the Laplace operator in terms of spherical coordinates our Poisson equation reads
    [tex]\frac{1}{r} \frac{\mathrm{d}^2}{\mathrm{d} r^2} (r \Phi)=-\frac{Q}{4 \pi R^2} \delta(r-R). \qquad (*)[/tex]
    For [itex]r \neq R[/itex] the right-hand side is 0 and the general solution thus reads
    with integration constants [itex]A[/itex] and [itex]B[/itex], which are to be chosen different for [itex]r<R[/itex] and [itex]r>R[/itex] due to the singularity of the [itex]\delta[/itex] distribution. We have to find the constants from appropriate boundary conditions.

    Start with [itex]r<R[/itex]. Since there are no sources or other singularities there, we must have [itex]A=0[/itex] in that region, because otherwise the potential would have a singularity in the origin. Thus we have
    [tex]\Phi(r)=B \quad \text{for} \quad r<R.[/tex]
    For [itex]r>R[/itex] we only demand that the potential vanishes at infinity. This fixes simply the overall additive constant of the potential, which has no physical meaning anyway.

    So our solution reads
    [tex]\Phi(r)=B \Theta(R-r)+\frac{A}{r} \Theta(r-R).[/tex]
    The potential should be continuous at [itex]r=R[/itex] since only its second derivative should have a [itex]\delta[/itex]-distribution like singularity. The first derivative must thus have a jump. So we must have [itex]B=A/R[/itex]. And thus
    [tex]\Phi(r)=A \left [\frac{1}{R} \Theta(R-r)+\frac{1}{r} \Theta(r-R) \right ].[/tex]

    To determine finally [itex]A[/itex], we have to multiply the equation (*) by [itex]r[/itex] and integrate over an infinitesimal interval around [itex]r=R[/itex]. This gives
    [tex]\left ([r \Phi(r)]' \right )_{r=R+0^+}-\left ([r \Phi(r)]' \right )_{r=R-0^+}=-\frac{Q}{4 \pi R}.[/tex]
    Evaluating the left-hand side gives
    [tex]-A/R=-\frac{Q}{4 \pi R} \; \Rightarrow \; A=\frac{Q}{4 \pi}.[/tex]
    The potential thus reads
    [tex]\Phi(r)=\frac{Q}{4 \pi} \left [\frac{\Theta(R-r)}{R}+\frac{\Theta(r-R)}{r} \right ].[/tex]
    Inside the spherical shell the potential is constant and thus the field vanishes there. Outside you have a Coulomb potential as if the charge is sitting at the center.

    Of course you can also evaluate the potential using the Green's-function method, according to which you just integrate over all points of your charge distribution, using Coulomb's Law, i.e.,
    [tex]\Phi(\vec{x})=\frac{1}{4 \pi} \int_{V} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}.[/tex]
    In your case the integral simplifies to an integral along the spherical surface,
    [tex]\Phi(\vec{x})=\frac{1}{4 \pi} \frac{Q}{4 \pi R^2} R^2 \int_0^{2 \pi} \mathrm{d} \varphi' \int_0^{\pi} \mathrm{d} \vartheta' \frac{\sin \vartheta'}{|R \vec{e}_r(\vartheta',\varphi')-\vec{x}|}.[/tex]
    Here I've used spherical coordinates [itex](\vartheta',\varphi')[/itex] to parametrize the spherical shell and set the radial unit vector to [itex]\vec{e}_r(\vartheta',var\phi')=(\cos \varphi' \sin \vartheta',\cos \varphi' \sin \vartheta',\cos \vartheta')[/itex]. Now, you can choose the spherical coordinates such that [itex]\vec{x}=r \vec{e}_z[/itex], i.e., that this vector points along the polar axis of the spherical coordinates. Then you have
    [tex]|R\vec{e}_r-\vec{x}|^2=(R^2 + r^2 - 2 R r \cos \vartheta').[/tex]
    The integral over [itex]\varphi'[/itex] thus just gives a factor [itex]2 \pi[/itex], and you get
    [tex]\Phi(\vec{x})=\Phi(r)=\frac{Q}{8 \pi} \int_0^{\pi} \mathrm{d} \vartheta' \frac{\sin \vartheta'}{\sqrt{R^2+r^2-2 R r \cos \vartheta'}}.[/tex]
    Now you substitute [itex]u=\cos \vartheta', \quad \mathrm{d} u=-\mathrm{d} \vartheta' \sin \vartheta'[/itex], which gives
    [tex]\Phi(r)=\frac{Q}{8 \pi} \int_{-1}^{1} \mathrm{d} \vartheta' \frac{1}{\sqrt{R^2+r^2-2 R r u}}.[/tex]
    This you can integrate in closed form, giving
    [tex]\Phi(r)=\frac{Q}{8 \pi} \frac{R+r-|R-r|}{r R} = \frac{Q}{4 \pi} \left [\frac{\Theta(R-r)}{R}+ \frac{\Theta(r-R)}{r} \right].[/tex]
    This is of course the same result as with the direct calculation of the Poisson equation.
  8. Apr 3, 2013 #7
    Thanks so much!! By the way is this approach to the Schroedinger equation using fractional derivatives good?

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    Last edited: Apr 4, 2013
  9. Apr 7, 2013 #8
    thanks very much!!! wow I really prefer your beautiful method.... as you know you can do this integral with single substitution or spherical coordinates also- here I integrate an annulus whose radius is a function of theta since there is azimuthal symmetry. maybe one can think of it as a double integral in phi and theta also

    [tex]\frac{1}{4\pi\epsilon o}\int_{0}^{\pi}\frac{2\pi Rsin\theta (Rd\theta)}{(R^2+z^2-2Rzcos\theta)^{3/2}}= \frac{\sigma R^2}{2\epsilon o}\int_{-1}^{1}\frac{z-Ru}{(R^2+z^2-2Rzu)^{3/2}}=[/tex]
    [tex]\frac{\sigma R^2}{2\epsilon o}\int_{|z-R|}^{|z+R|} \frac{(z-R(\frac{z^2+R^2-y^2}{2zR}))dy}{zRy^{2}}=\frac{\sigma R}{4z^2\epsilon o}\int_{|z-R|}^{|z+R|} \frac{(2z^2-z^2-R^2+y^2)dy}{y^2}=[/tex]
    [tex]\frac{\sigma R}{4z^2\epsilon o}\int_{|z-R|}^{|z+R|} \frac{(z^2-R^2+y^2)dy}{y^2}=\frac{\sigma R}{4z^2\epsilon o}[\frac{-(z^2-R^2)}{y}+y]_{|z-R|}^{|z+R|}=\frac{\sigma 2\pi R^2}{4\pi z^2\epsilon o} [\frac{z-R}{|z-R|}+\frac{z+R}{|z+R|} ][/tex]
    Last edited: Apr 7, 2013
  10. Apr 26, 2013 #9
    it's interesting how Gauss's law can also be written in differential form using the Dirac delta function and there is a gravity form

    [tex]\bigtriangledown \cdot E=\frac{1}{4\pi \varepsilon o}\int 4\pi \delta ^3(r-r')\rho (r')d\tau '=\frac{Q}{\varepsilon o}[/tex]

    as you know here is a proof for the uniqueness theorem for Poisson's equation
  11. Apr 30, 2013 #10
    Last edited by a moderator: Sep 25, 2014
  12. Jun 9, 2013 #11
    Thanks so much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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