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The following statement is from the book Concepts of Physics by Dr. H.C.Verma, from the chapter on "Capacitors", under the topic "Spherical Capacitor - Isolated sphere":

enter image description here

If we assume that the outer sphere [$A$] is at infinity, we get an isolated single sphere of radius $R_1$. The capacitance of such a single sphere can be obtained from equation $(31.3)$ by taking the limit as $R_2\to\infty.$ Then

$$C=\frac{4\pi \epsilon_0 R_1R_2}{R_2-R_1}\tag{31.3}$$

$$\approx \frac{4\pi \epsilon_0 R_1R_2}{R_2}=4\pi\epsilon_0R_1$$

If a charge $Q$ is placed on this sphere, its potential (with zero potential at infinity) becomes

$$V=\frac Q C = \frac{Q}{4\pi\epsilon_0R_1}$$

(Emphasis mine)

I understood the given situation mathematically. But, even after assuming the charged sphere $A$ extends till infinity, how do we still consider zero potential (reference point for electric potential) at infinity? I think the potential at infinity is no longer zero due to the presence of the charged sphere $A$ there and hence anything based on the assumption "potential at infinity is zero" must fail. But the final result obtained is correct even if we neglect this fact? How is this possible?

Image constructed by me with the help of diagram 31.5 from the book mentioned.

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A potential can always by changed by adding an arbitrary constant, and this has no effect on the physics. In this example, you can choose the potential to be zero at any radius you like. You could choose it to be zero at the surface of the sphere, for instance. But that would make the equation look a little more complicated, so we usually don't do that. (You could also add a constant with the same sign as $Q$, so that the potential would never be zero.)

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  • $\begingroup$ Thank you. After reading your answer, I realised that when the radius of a spherical shell having a finite charge density, is gradually increased to infinity, the surface charge density gradually decreases to zero. And I think it'll also be consistent with the existing choice of zero potential. $\endgroup$
    – Vishnu
    Dec 17, 2019 at 13:39

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