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I am doing a question on capacitance and coulombs. I got the answer correct, but I was wondering how a physicist, when doing the calculation on paper, would differentiate a C and C?

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closed as primarily opinion-based by Kyle Kanos, ZeroTheHero, Jon Custer, Void, Norbert Schuch Jan 8 at 0:25

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    $\begingroup$ Capacitance is measured in Farads, with an "F". $\endgroup$ – Hot Licks Jan 5 at 2:05
  • $\begingroup$ When the equation is in terms of symbols, $C$ probably indicates some capacitance. When you are using the equation to calculate something by substituting numerical values for all the symbols, then C is probably the unit (Coulomb) for the value of some symbol that represented a charge. This does not tend to create confusion unless you substitute values for some symbols and not others. $\endgroup$ – G. Smith Jan 5 at 3:22
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I will omit the fact that capacitance is measured in Farads, as I see a more general use to the question. For the sake of this answer I will answer as if we are symbolically manipulating capacitance and use the variable $C$.


I usually don't have this issue with capacitance, but when I want to make explicit that I'm using a capital letter that is non italicized what I'll do is put a line under it. For instance I would write $C$ for capacitance and $\underline{C}$ for coulombs.

Like I said this is probably overkill for capacitance and coulombs, but I think it's useful in a broader context.

Downfalls: If you're writing fast it can get kind of messy if your capital letter appears in the numerator like $\frac{\underline{C}}{C}$ and wind up looking awkward, so I only do this if I know it's not going to be an issue.

2$^{nd}$ solution:If it is unavoidable and the algebra will get nasty I will write my capital letters as "double struck". So for instance $\mathbb{C}$ would be Coulombs in this case, and I just close my eyes and hope that I don't need to mention the complex numbers.

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    $\begingroup$ Or just C for coulombs, C1, C2, ... for different capacitances. $\endgroup$ – The Photon Jan 5 at 1:35
  • $\begingroup$ That's fair, but I was trying to give an answer that was more general and so may be useful ini a broader context $\endgroup$ – InertialObserver Jan 5 at 1:36

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