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I know $\mathrm{J} = \mathrm{N\ m} = \mathrm{C\ V}$, but what if I want joules expressed in SI base units of $\mathrm{kg}$, $\mathrm{m}$, and $\mathrm{s}$? This is trivially easy if I use $\mathrm{N\ m}$ to get there: $$\mathrm{N} = \mathrm{kg\ m/s^2}$$ ergo using only SI base units to derive $\mathrm{J}$ we get $$\mathrm{J} = \mathrm{kg\ m^2/s^2}$$

But if the only equation I had was $\mathrm{J}=\mathrm{C\ V}$, how does one get from $\mathrm{C\ V}$ to $\mathrm{kg m^2/s^2}$?

It seems to me clear that $\mathrm{C}$ is the electrical analog of $\mathrm{kg}$, and $\mathrm{V}$ (electromotive force) would be accelerating the charge just as $G$ (force of gravity on earth) would be accelerating the mass. But what trips me up is that charge and acceleration are the only 2 components of the electrical equation ($\mathrm{J} = \mathrm{CV}$) but in the mechanical equation ($\mathrm{J} = \mathrm{kg\ (m/s^2)\times m}$) there is mass, acceleration and distance. So there seems to be a missing term and so conceptually $\mathrm{C\ V}$ cant equal $\mathrm{N\ m}$ nor be expressed by $\mathrm{kg m^2/s^2}$.

Of course I know they are equivalent. I just cant draw the line from point A to point B. Do I have to do something nutty and implausible like convert to Gaussian Units and then convert back to SI to get there?

I am a total amateur trying to teach myself , besides the actual unit conversions from $\mathrm{N\ m}$ to $\mathrm{C\ V}$ (assuming such conversion is possible), if this can be explained at a conceptual level and/or as rudimentarily as possible that would be great (for example creating analogs between the gravitational and electrical worlds (like am I correct that charge an analog of mass, is there any gravitational analog to I (current), if not why not, is voltage an analog for acceleration due to gravity? etc). Thanks!

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  • $\begingroup$ Is there a reason you did not just write V and C in terms of the base units (like you did with J) and simplify? Why all of the analogy stuff when the direct calculation is so easy? $\endgroup$
    – Dale
    Commented May 24, 2020 at 12:38
  • $\begingroup$ How would one do that? I see the steps From Nm to SI base units. W= Fd. F =ma, a = v/t, and v = d/t so W = mdd /tt (kg m^2/s^2). Easy. But can you show me the easy way to express C and V in terms of m or a since I know of no equation that expresses m or a as part of C or V. IOW, I didn't write V or C in terms of kg etc cuz I don't know how to simplify CV to kg m s as I just did with Nm. What are the equations that show kg or m or s as a component of C or V? Can you give an example? Or explain it like I'm a kid, or just "write V and C in terms of base units and simplify"so I can see it done $\endgroup$
    – A Anderson
    Commented May 24, 2020 at 18:15
  • $\begingroup$ The SI base units are listed on their wiki pages: en.m.wikipedia.org/wiki/Volt and en.m.wikipedia.org/wiki/Coulomb $\endgroup$
    – Dale
    Commented May 24, 2020 at 19:11
  • $\begingroup$ Dale thanks for your time. I had spent alot of time on the wikipedia entries for volts and coulomb to no avail, but found an article called The Relation of Ohm’s Law to Newton’s 2nd Law. It shows "If charge is measured as a wave amplitude, which is a distance, then all of the units align and the equations can be consolidated" So the mechanical analog for charge is DISTANCE not mass. If your interested here is a link .pdfs.semanticscholar.org/7232/… Thanks again. $\endgroup$
    – A Anderson
    Commented Jun 5, 2020 at 1:15

2 Answers 2

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Analogies can give useful insights, but they have to be treated with care. I'm afraid that your analogy is a little too loose...

(1) One problem is that in Newtonian Physics, mass has a dual role: it not only 'feels' the pull of gravity, but it reduces the acceleration that a given force will give the body (the inertial role of mass). [That's why heavy bodies fall with the same acceleration as light bodies, if there is no air resistance.] Charge does not have this dual role.

(2) In J = CV, V is the unit of potential difference. Potential difference is, essentially, force-per-unit-charge multiplied by distance. So, in the terms of your question, the distance factor is contained in the V.

Hope this helps. But I'm afraid there's no substitute for using precise arguments based on precise definitions.

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  • $\begingroup$ Thanks very helpful! What also helped was a short and easily understood paper called The Relation of Ohm’s Law to Newton’s 2nd Law. It shows "If charge is measured as a wave amplitude, which is a distance, then all of the units align and the equations can be consolidated" So the mechanical analog for charge is DISTANCE not mass. Jeff Yee, the author, then uses charge (measure in meters) to create mechanical analogs for all the other electrical entities. If your interested here is a link .pdfs.semanticscholar.org/7232/… $\endgroup$
    – A Anderson
    Commented Jun 5, 2020 at 1:08
  • $\begingroup$ Thank you for your comment and for the link. If you're interested in understanding Physics, I would advise you to ignore the contents of the link. $\endgroup$ Commented Jun 5, 2020 at 7:48
  • $\begingroup$ Really? Why? Maybe I'm trying to hard too find an analog, but it seems that, at some level, there must be a way to reconcile how Nm = CV. I thot the paper did so, but now I am curious why not. I also wonder if maybe there is no good analog between electric and gravitational fields cuz only one is polar. But it still seems to me that a force (from any field) acts similarly to objects "feeling" that form of field energy. Sorry for my ignorance. but I genuinely am confused. I can see how distance would be contained in voltage (just as gravitation potential is tied to distance) or in charge. $\endgroup$
    – A Anderson
    Commented Jun 5, 2020 at 13:23
  • $\begingroup$ Oh I think I get it. J = CV means V = J/C = Nm/C (as you said in your original answer "force per unit charge", which only half sunk in). So I was still equating voltage to the gravitational force. But it is like gravity per unit of _____ (mass I suppose). I hope that is correct. If so thanks again, if not, back to the drawing board for me $\endgroup$
    – A Anderson
    Commented Jun 5, 2020 at 13:38
  • $\begingroup$ @ A Anderson Good; I'm glad you're getting it. Regarding Mr Yee's paper, measuring charge in metres isn't justified and leads to confusion. I'm sorry to be boring, but if you want to understand Physics, get a good textbook and work through it, doing the exercises and learning the definitions, laws and derivations. By all means consider analogies, but let them be firmly grounded in the definitions of the quantities involved. Good luck! $\endgroup$ Commented Jun 5, 2020 at 14:17
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I'm not sure of the exact nature of your confusion here, but one thing that stands out is that it sounds like you might be mixing up units (joules, coulombs, volts, newtons, meters) with the physical quantities they represent (energy, charge, electrical potential, force, distance). You spent quite a bit of this question asking about analogs between electrical and mechanical systems, and while it might make sense to do that when talking about certain physical quantities, it doesn't make nearly as much sense with units. Different types of quantities are measured by different units, and that's it. There's no particular reason to expect those units to behave similarly.

For instance, a coulomb is defined as a unit of electrical charge. A volt is defined as a unit of a quantity which is equivalent to an amount of energy per unit charge. So if you multiply a coulomb by a volt, you get $$\text{charge}\times\frac{\text{energy}}{\text{charge}} = \text{energy}$$ which is, of course, measured in joules. Note that you don't have to think about which charge, or what exactly it means to have an amount of energy per unit charge, or what else might be equivalent to that. Unit definitions are very simple algebra.

Now, to be precise, what I just wrote doesn't necessarily tell you that $CV = J$, but it does tell you that $CV \propto J$; that is, you might have $CV = J$ or $CV = 2J$ or $CV = \frac{1}{1000}J$ or so on. The number is determined by how all the various units are defined. Part of what makes the SI standard so convenient is that they've chosen definitions of the different units so that many of these factors are equal to one.

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  • $\begingroup$ David, thank you. Also very helpful. If your interested I found a paper that cleared up my struggle trying to find analogs, called The Relation of Ohm’s Law to Newton’s 2nd Law. It shows "If charge is measured as a wave amplitude, which is a distance, then all of the units align and the equations can be consolidated" So the mechanical analog for charge is DISTANCE not mass. Jeff Yee, the author, then uses charge (measured in meters) to create mechanical analogs for all the electrical entities. Here's a link .pdfs.semanticscholar.org/7232/… $\endgroup$
    – A Anderson
    Commented Jun 5, 2020 at 1:12

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