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Everything I saw suggests coulomb has no fundamental units. So then how is Newton/Coulomb equivalent to Volt/meter?

DanielSank
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most venerable sir
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3 Answers3

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The coulomb (C) is the derived unit for charge in SI. One coulomb is the amount of charge in one ampere-second. The elementary charge (charge of one proton or (-) electron) is roughly ${\sim}1.602\times10^{-19}\:\mathrm C.$ Moreover, $1\:\mathrm C =1\:\mathrm{A\,s}$.

The units work out the in the equivalency, which never requires breaking the coulomb into its base units, as follows:

  • Newton: amount of force used when accelerating one kilogram at 1 meter per second squared, $$1\:\mathrm N = 1\mathrm{\frac{kg \, m}{s^2}}.$$

  • Joule: energy transferred (or work done) when applying a force of one newton through a distance of one meter, $$1\:\mathrm J =1\:\mathrm{N \, m}.$$

  • Volt: potential energy of one joule per electric charge of one coulomb, $$1\:\mathrm V = 1\:\mathrm{\frac{J}{C}}.$$

  • Converting volts to joules per coulomb, then joules to newton-meters, then cancelling the meter, you get $$\mathrm{ 1\:\frac{N}{C} = 1\:\frac{V}{m} = 1\:\frac{J}{C \, m} = 1\:\frac{N \, m}{C \, m} = 1\:\frac{N}{C}. }$$

Emilio Pisanty
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usncahill
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    AFAIK, there is no such thing as a "fundamental unit" defined in SI. There are base units and derived units. The coulomb is a derived unit. The ampere and the second are base units, and 1 coulomb is equal to 1 amp-second. For more info, see the NIST Guide for the use of SI. – The Photon Apr 23 '15 at 02:01
  • Ah! Thanks! Got amp and coulomb swapped in my head. Mind if I change my answer to match, or are you posting one? – usncahill Apr 23 '15 at 02:25
  • For the record, current as a base unit is pretty unintuitive. – usncahill Apr 23 '15 at 02:35
  • It is unintuitive. The questions I linked in commments on the OP ask about it explicitly. – The Photon Apr 23 '15 at 02:36
  • Ah! Makes sense. What's a kg; the weight of that ball over there. What's an ampere; the current moving thru this. What's a Coulomb; well, count the number of particles moving thru the wire ... haha. Thanks for the links. – usncahill Apr 23 '15 at 02:38
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The SI units for electromagnetism are based on the ampere, which is

that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to $2\times10^{-7}$ newton per meter of length.

One coulomb is the amount of charge delivered by a current of one ampere when integrated over one second ($\rm 1\,C = 1\,A\cdot s$).

One volt is the electric potential difference across a device which consumes one watt of electrical energy when powered by a current of one ampere ($\rm 1\,V = 1\,\frac WA$). This means that the volt is also equal to the electrical potential energy of one coulomb of charge ($\rm 1\,V = 1\,\frac{W\cdot s}{A\cdot s} = 1\,\frac JC$).

In your example, $$ \rm 1\,\frac NC = 1\,\frac NC \frac mm = 1\,\frac JC \frac 1m = 1\,\frac Vm. $$

Note that beginning in 2018 the SI units will be overhauled to depend on fundamental constants. After the redefinition the coulomb will be defined as "the negative of the charge of $\frac{10^{19}}{1.602\cdots}$ electrons", where the first eight or nine significant digits of the denominator are already known.

rob
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E=V/r=volt/meter as volt=Joule/coulomb E=J/C.m as J=N.m E=N.m/C.m as m and m cancel each other so,E=N/C

momin
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