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In page-87 of Purcell, David J. Morin Electricity and magnetism (3 ed) , the following theorem is discussed:

Theorem 2.1 If $\phi(x, y, z)$ satisfies Laplace’s equation, then the average value of $\phi$ over the surface of any sphere (not necessarily a small sphere) is equal to the value of $\phi$ at the centre of the sphere.

The proof starts with bringing charge ($q'$) from infinity to a sphere(of charge $q$) which was easy for me to understand, the second part involves bringing the sphere to the charge which I find difficult to understand. I quote the second part:

Now suppose, instead, that the point charge $q$ was there first and the charged sphere was later brought in from infinity. The work required for that is the product of $ q′$ and the average over the surface $S$ of the potential due to the point charge $q$. Now the work is surely the same in the second case, namely $\frac{qq′}{4π\epsilon_0R}$, so the average over the sphere of the potential due to $q$ must be $\frac{q}{4 \pi \epsilon_o r}$. That is indeed the potential at the centre of the sphere due to the external point charge $q$.

I have two main questions on the above paragraph:

  1. What exactly does it mean to average a function over a surface?
  2. Why should the work be the product of the charge $q'$ with the average potential?
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1 Answer 1

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  1. To average a function $f(\mathbf r)$ over a sphere means to integrate the value of the function over the sphere and divide by the area of the sphere. $$\frac{1}{4\pi R^2}\int_{\rm sphere}f(\mathbf r)\,dS.$$ This is the usual way to define averages.

  2. If we bring a point charge $q'$ to a point in space where there is a potential $V(\mathbf r)$ created by other charges, then the work needed is $q'V(\mathbf r)$. If, as in your case, the charge that we bring is not a point charge, but is distributed over a sphere of radius $R$, we need to sum (integrate) the work done to bring each infinitesimal charge at each point in the sphere $dq=\sigma dS$ multiplied by the potential at that point in the sphere: $$W=\int_{\rm sphere} V(\mathbf r)dq=\int_{\rm sphere} \sigma V(\mathbf r) dS.$$ Since the total charge in the sphere is $q'$, then $\sigma=q'/(4\pi R^2)$, so that the work needed is $$\int_{\rm sphere} \sigma V(\mathbf r) dS=q'\underbrace{\left[\frac{1}{4\pi R^2}\int_{\rm sphere} V(\mathbf r) dS\right]}_{\text{average of V over the sphere}}.$$

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  • $\begingroup$ The integral looks kinda sketch because as you bring in more and more charge, then the charges on sphere would cause a work of it's own. So the total work would have to include that as well (not just the work against pushing charge w.r.t the fixed charge which is kept originally) $\endgroup$
    – Babu
    Commented Apr 27, 2021 at 18:35
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    $\begingroup$ Basically, how are you accounting for the changing charge density ? $\endgroup$
    – Babu
    Commented Apr 27, 2021 at 18:36

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