How do I find the flux through a sphere which is off centre with respect to a charge?

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Solution 1

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Let a spherical surface of radius $\,R\,$ with center at point $\,O\,$ and a positive charge $\,Q\,$ eccentrically placed inside the sphere at a distance $\,b<R\,$ from the center $\,O$, as in above Figure(1). We'll try to find the flux of the electric field intensity $\,\mathbf{E}\,$ through this surface by elementary calculus without use of Gauss' Law or divergence theorem.

Since we have rotational symmetry around the axis $\,OQ\,$ we cut an infinitesimal ring $\,\mathrm{dS}\,$ as in Figure. This ring has perimeter $2\pi R\sin\phi=2\pi\; r\sin\theta$ and width $R\, \mathrm{d}\phi$, so infinitesimal area \begin{equation} \mathrm{dS}=2\pi R^{2}\sin\phi\, \mathrm{d}\phi = 2\pi R\, r \sin\theta\, \mathrm{d}\phi \tag{01} \end{equation} The electric field intensity $\,\mathbf{E}\,$ is at any point of $\,\mathrm{dS}\,$ collinear with $\,QP\,$, inclined in general from the normal by an angle $\,\omega=\theta-\phi\,$, see in the Figure. Flux is produced by its normal component $\,\mathbf{E}_{\boldsymbol{\perp}}\,$ only. Note that the magnitude of $\,\mathbf{E}\,$ and the angle $\,\omega$ are constants on the infinitesimal ring $\,\mathrm{dS}\,$ and so the magnitude of the normal component $\,\mathbf{E}_{\boldsymbol{\perp}}\,$ \begin{align} \mathrm{E} & =\Vert\mathbf{E}\Vert=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{r^{2}} \tag{02}\\ \mathrm{E}_{\boldsymbol{\perp}} & =\Vert\mathbf{E}_{\boldsymbol{\perp}}\Vert=\mathrm{E}\cos\omega=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{r^{2}}\cos\omega \tag{03} \end{align} For the infinitesimal flux $\,\mathrm{d\Phi}\,$ we have from equations (01) and (03) \begin{equation} \mathrm{d\Phi}=\mathrm{E}_{\boldsymbol{\perp}}\cdot\mathrm{dS}=\left(\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{r^{2}}\cos\omega\right)\cdot\left(2\pi R^{2}\sin\phi\, \mathrm{d}\phi\right) \tag{04} \end{equation} that is \begin{equation} \mathrm{d\Phi}=\frac12\dfrac{Q R^{2}}{\epsilon_{0}}\dfrac{\cos\omega}{r^{2}}\sin\phi\, \mathrm{d}\phi \tag{05} \end{equation} Now, from triangle $\,OQP\,$ we have \begin{align} \cos\omega & =\dfrac{R^{2}+r^{2}-b^{2}}{2Rr} \tag{06}\\ \cos\phi & =\dfrac{R^{2}+b^{2}-r^{2}}{2Rb} \Longrightarrow \sin\phi\, \mathrm{d}\phi = \dfrac{1}{R\,b}\, r\, \mathrm{d}r \tag{07} \end{align} Replacing these expressions of $\,\cos\omega\,$ and $\,\sin\phi\, \mathrm{d}\phi\,$ in (05) we have \begin{equation} \mathrm{d\Phi}=\dfrac{1}{4b}\dfrac{Q}{\epsilon_{0}}\left(\!1\!+\!\dfrac{R^{2}-b^{2}}{r^{2}}\right)\mathrm{d}r \tag{08} \end{equation} and integrating to include all the spherical surface \begin{equation} \Phi = \int\limits_{r=R-b \vphantom{\frac12}}^{r=R+b\vphantom{\frac12}}\!\!\!\!\!\!\mathrm{d\Phi}=\dfrac{1}{4b}\dfrac{Q}{\epsilon_{0}}\underbrace{\left[\:\:\int\limits_{r=R-b \vphantom{\frac12}}^{r=R+b\vphantom{\frac12}}\!\!\!\!\!\!\left(\!1\!+\!\dfrac{R^{2}-b^{2}}{r^{2}}\right)\mathrm{d}r\right]}_{=4b}=\dfrac{Q}{\epsilon_{0}} \tag{09} \end{equation}


(1) Appended a Figure and its 3D version. The charge $Q$ is on place $\xi$.

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Solution 2

Step 1: Draw a small sphere within the large sphere and centred on the charge. Now it's easy to calculate the flux through the small sphere and you can prove Gauss' law for the small sphere.

Step 2: Every field line that passes through the small sphere must also pass through the large sphere on its way out to infinity. Therefore the total flux through the large sphere must be the same as the flux through the small sphere. This proves Gauss' law for the large sphere as well.

It is of course possible to do the proof by brute force if you enjoy that sort of calculational exercise.

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Nigel Goveas
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Updated on August 07, 2020

Comments

  • Nigel Goveas
    Nigel Goveas over 3 years

    Suppose I have a charge $Q$ at the origin and a sphere with equation $$(x-a)^2+(y-b)^2+z^2=r^2$$ So the electric field is not normal to spherical surface in general.

    How do I prove that the flux through the sphere is $\dfrac{Q}{\epsilon}$?

    • Nigel Goveas
      Nigel Goveas over 6 years
      How did you come up with this conclusion?
    • Nigel Goveas
      Nigel Goveas over 6 years
      Actually I wanted a proof of not the r²<a²+b² lol. I wanted the mathematical proof of my question with the charge being inside the sphere
    • Nigel Goveas
      Nigel Goveas over 6 years
  • Michael Seifert
    Michael Seifert over 6 years
    Nicely done. If nothing else, doing a calculation this way can help you realize how powerful Gauss's Law actually is.
  • Nigel Goveas
    Nigel Goveas over 6 years
    I thought you would use solid angle method but boy you're way above it. Can you also tell me what software you've used to get those pictures? I'm so indebted to you
  • Michael Seifert
    Michael Seifert over 6 years
    @NigelGoveas: The idea here is to do a change of variables to write the whole flux integral as an integral over $r$, the distance from the point charge to the surface. The transformation in equation (07) is just the usual equation for change of variables in a single-variable integral (i.e., "$u$-substitution".)
  • Nigel Goveas
    Nigel Goveas over 6 years
    @Michael Seifert- So we only consider that for a small area called ds only Φ and r are varying.Correct me if I'm wrong but R and we are also varying here but I'm guessing the limits of integration takes care of them. Am I right?