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Nov 23, 2023 at 3:51 comment added knzhou Certainly, you can think of it that way!
Nov 23, 2023 at 2:54 comment added Jonathan Huang So the problem isn't that the charge is not conserved, but that what we fixed at the start wasn't the charge, but the potential $V_1$, and the series of approximations tells us the charge $Q^{(1)}$ that is needed to produce such a potential?
Nov 23, 2023 at 2:51 comment added knzhou For example, suppose I was trying to find increasingly good approximations to the energy of a particle. The leading term is $E_0 = mc^2$. The first correction is $E_1 = mv^2/2$. The total answer is $E = E_0 + E_1 + \ldots = mc^2/\sqrt{1-v^2/c^2}$. Your question is analogous to saying: "Wait a second! $E$ is bigger than $E_0$! Doesn't your argument violate conservation of energy?"
Nov 23, 2023 at 2:47 comment added knzhou Smythe is just trying to find a self-consistent solution to the equations. The sum of all the $Q_n$ is the physical charge on the sphere. The quantity $Q_1$ has no physical meaning. It is just some auxiliary quantity that you define early on in the argument, but only the result of the full argument has meaning.
Nov 23, 2023 at 2:11 comment added Jonathan Huang @Triatticus $Q^{(1)}_n$ is always positive, so the sum of the series must be larger than $Q^{(1)}_1$.
Nov 23, 2023 at 2:10 history edited Jonathan Huang CC BY-SA 4.0
grammar nuissance
Nov 23, 2023 at 2:10 comment added Jonathan Huang @LPZ yes, that is what I mean. I edited the sentence.
Nov 22, 2023 at 23:47 comment added LPZ What do you mean by the charge you started with ? Did you mean write in the last equation: $$Q^{(1)}=Q_1^{(1)}$$ if not, what is $Q$?
Nov 22, 2023 at 22:23 comment added Triatticus How did you come to the conclusion that the charge is more? Did you explicitly evaluate that series?
Nov 22, 2023 at 18:20 history asked Jonathan Huang CC BY-SA 4.0