I made some calculations and I got a correct answer, although I lack the intuition and I have no idea why it works. Say we want to build a uniformly charged sphere with radius R and total charge Q, bringing charges from infinity and building it. In order to construct the shell, we need to do work against the force acting on every infinitesimal area element. The electric field outside the shell is $E=\frac{kQ}{r^2}$ and inside is zero. Therefore, we get:

$$p=\frac{F}{A}=\frac{1}{2}\epsilon_0 (E_1-E_2)^2 = \frac{1}{2}\epsilon_0\left(\frac{k^2Q^2}{r^4}\right)$$

and integrating we get:

$$W=\int_R^\infty\int_0^{4\pi r^2}\frac{1}{2}\epsilon_0\left(\frac{k^2Q^2}{r^4}\right)\, dadr\, . $$

The result is the work done to construct only the shell. Now, what about the "inside" of the sphere? I did the same, but with $E=\frac{kQr}{R^2}$ and the radius ranging from zero to R and summed the two works (building the shell + building the inside) and I got the right result.

Now, my question is, why? For calculating the work done in order to build the inside of the sphere, I used the electric field inside a uniformly charged sphere, but what we actually do is build infinitely many shells, so I don't see how that works. I will be very glad for some clarification and intuition.

*I know there are easier ways to calculate it.

  • $\begingroup$ Another way to calculate it(more intuitive) would be assume that you have already "made" a sphere of a smaller size, of charge q and radius r. As a dq charge is brought from ∞ to the surface, the small work done Is potential on the surface of the sphere dW= (Kq/r) *dq by definition. Now charge density = Q/(4/3πR³) = q/(4/3πr³) or Q/R³ = q/r³, now express r in terms of q,Q,R and integrate the expression you'll have your answer $\endgroup$ – user600016 Apr 22 at 2:01

If you're OK with Riemann sums as an intuition for the integral in general (say, integrating $y = x$ over $x$ from $0\rightarrow 1$), then the intuition behind the process of building a sphere with a bunch of infinitesimal shells should be surmountable. I don't think you're saying this is the problem, but I'm saying it anyway to keep the analogy alive between infinitesimal shells and infinitesimally width'd rectangles. Not to say this is the only intuition for integration ;)

It seems to me that you feel the weirdness is coming from the order in which the parts of the sphere are constructed, i.e., how could the inside of the sphere be created after the outside? How can I claim the outside has been constructed before constructing it!? To this, I'd say because the electrostatic field is defined as conservative (in the scope of electrostatics this is most simply seen with the ideas that (1) $\nabla \times \vec{E} = 0$ and (2) $\vec{E} = -\nabla V$), you can generate the parts of the sphere in any order and the answer would be correct.

I.e., the exact process taken to create the uniformly charged sphere doesn't matter.

In the same way that you add on at the end that you "know there are easier ways to calculate it", you are able to construct the sphere however you wish--so long as the final result is the same--the way to get to the result doesn't matter.

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  • $\begingroup$ Yes, my problem is exactly with the order of building the sphere. I still don't get why would we use the electric field inside a charged sphere when there isn't a full "ball" while constructing it, or how the process would coherently look like? I try imagining the process but I fail. $\endgroup$ – Darkenin Apr 21 at 19:42
  • $\begingroup$ I think you're too focused on the casualty implied when someone says that the integral is 'building the sphere'--in that building implies one thing happens before another and possibly that the order is important. My answer above tries to relay that there is no importance to the order. And moreover, there doesn't need to be any intuition, it just so happens that some intuition exists when building from $r:0\rightarrow R$. Breaking that toy model for intuition is having the obvious result ;) $\endgroup$ – outyprouty Apr 22 at 20:28

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