# How can we consider charge to be continuous? [duplicate]

In electrostatics, we usually consider charge to be continuous on any body, to calculate the electric field of the body. For eg. I had proved the Shell Theorem taking an infinitesimal charge of $dq$ on the sphere. But we also know that charges are quantized (In terms of electrons and protons). They are finite and not infinitesimal. So is considering them continuous an approximation or is there a reason to why we can do it?

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## marked as duplicate by Dan, Qmechanic♦Jul 10 '13 at 23:37

Considering the charge to be continuous is an approximation, but since the electron charge is $1.60217657 \times 10^{-19}$ coulombs it's a very good approximation. – John Rennie Jul 9 '13 at 14:42
Yes, the continuity of the electric charge is an approximation that is valid whenever the relevant charges are much greater than the elementary charge (of the electron, or the proton). When we deal with numbers like $1,234,567\,e$, it doesn't really matter that it should have been $1,234,567.8\, e$: very large numbers may be approximated by a nearby integer while the error is small.
This approximation is perfectly OK for macroscopic circuits. For example, if the currents are of order 1 Ampere, then it means that 1 Coulomb goes through the wire each second. But one Coulomb is about $10^{19}$ elementary charges, a number much greater than one, so the relative error is just $10^{-19}$ or so if we approximate the number of elementary charges by the nearest integer.