Question

A spherical cloud of charge of radius R carries total charge Q. The charge is
distributed so that its density is spherically symmetric, i.e. it is a function of the
radial distance from the centre of the sphere.
Explain why the “charge cloud” is equivalent to a point charge of Q Coulombs
at the centre of the sphere.

Answer 1

That is a tough problem -- or an easy one, depending upon the toolkit you bring to bear upon it. You may wish to consult David Griffiths book "Introduction to Electrodynamics", 3rd edition. The problem -- for uniform charge density on a spherical shell -- shows up on page 64, as problem 2.7, to be done as a direct use of Coulomb's law (very hard work). And on pp 70-71, as example 2.2 for a uniformly charged solid sphere, as an application of Gauss's law (fairly easy work). Which illustrates how useful Gauss's law can be.

The spherical shell, and the spherically symmetric cloud in your original problem, are really the same problem. In the more general form you state the problem, imagine the case of the charge being zero except in the neighbourhood of a certain radius R -- that is the same as a spherical shell of radius R. So if you have the solution to your problem as stated, you get the spherical shell result. And on the other hand, if you have the result for a spherical shell, you can add up (integrate) thin shells each with distinct charges to get your result.

What I'll try to do is convince you that the result is reasonable. Bit of arm-waving here. Imagine a spherical shell with uniform charge density on the shell. By symmetry, the force (due to electric field acting on some test charge) must act along a radium from the centre of the sphere. Suppose the shell is radius R and distance r from the test charge to the centre of the shell. (r > R). There is an electric field which can be written as (constant 1/4*pi*e0) times Q (charge on sphere) times some number 1/s^2, where s is the "effective distance" from the test charge. Compare s to r, the distance from the test charge to the centre of the sphere. If s = r, always, the desired result holds.

Suppose s does not equal r from some direction of the test charge, let's say s < r. Imagine the same test charge on the other side of the sphere, distance r from the centre. Everything is symmetric, and so the effective distance s2 from the other-side test charge must be equal to s. Furthermore, s2 < r by the symmetry. But that is impossible, because s + s2 = 2*r. Hence s=r.

Hope that helps you visualize the physicality of the situation. Not exactly a proof, but has the key idea. See Griffiths book for the more detailed approach, and for solution as an application of Gauss's law.

Another resource is the Yale online course, "Fundamentals of Physics II", Physics 201, taught by Prof Ramamurti Shankar.
http://oyc.yale.edu/physics/fundamentals-of-physics-ii
I think what you want is in lecture 3. Those are really excellent lectures, and there are transcripts of the spoken portions, so you can locate the content you want. The transcripts do not have the formulas -- you really have to watch the lecture videos to get Shankar's meaning -- but at least with the transcripts, you can confirm which lecture covers the material you are interested in.

all the above is just my present understanding. I'm also a student of this material. Anything you think is incorrect in what I've written, probably indeed is incorrect! I'll look forward to other people's replies on this topic as well, and anything else in electricity and magnetism.

Best wishes!

Answer 2

Radius, not "radium"!