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16.2: The CGS Electrostatic System

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  • Definition. One CGS esu of charge (also known as the statcoulomb) is that charge which, if placed 1 cm from a similar charge in vacuo, will repel it with a force of 1 dyne.

    The following exercises will be instructive.

    Example \(\PageIndex{1}\)

    Calculate, from the SI electricity that you already know, the force between two coulombs placed 1 cm from each other. From this, calculate how many CGS esu of charge there are in a coulomb. (I make it 1 coulomb = \(2.998 \times 10^9\) esu. It will not escape your notice that this number is ten times the speed of light expressed in \(\text{m s}^{-1}\).) Calculate the magnitude of the electronic charge in CGS esu. (I make it \(4.8 \times 10^{-10}\) esu. If, for example, you see that the potential energy of two electrons at a distance \(r\) apart is \(e^2/r\), this is the number you must substitute for \(e\). If you then express \(r\) in cm, the energy will be in ergs.)

    Coulomb's law in vacuo is


    This differs from our accustomed formula (equation 16.1.2) in two ways. In the first place, we are using an unrationalized definition of permittivity, so that the familiar \(4\pi\) is absent. Secondly, we are choosing units such that \(4\pi \epsilon _0\) has the numerical value 1, and so we are omitting it from the equation.

    While some readers (and myself!) will object, and say that equation 16.2.1 does not balance dimensionally, and is valid only if the quantities are expressed in particular units, others will happily say that equation 16.2.1 shows that the dimensions of \(Q\) are \(\text{M}^{1/2}\text{L}^{3/1}\text{T}^{-1}\), and will not mind these extraordinary dimensions.

    Electric field \(\textbf{E}\) is defined in the usual way, i.e. \(\textbf{F}=Q\textbf{E}\), so that if the force on 1 esu of charge is 1 dyne, the field strength is 1 esu of electric field. (The esu of electric field has no name other than esu.) This is fine, but have you any idea how this is related to the SI unit of \(\textbf{E}\), volt per metre? It requires a great deal of mental gymnastics to find out, so I'll just give the answer here, namely

    \[\nonumber 1 \text{CGS esu of} \ \textbf{E} = 10^{-6} c \text{V m}^{-1},\]

    where \[c=2.997 924 58 \times 10^{10}.\]

    Now the vector \(\textbf{D}\) is defined by \(\textbf{D} = k\textbf{E}\), and since the permittivity is held to be a dimensionless number,\(\textbf{D}\) and \(\textbf{E}\) are held to be dimensionally similar (\(\text{M}^{1/2}\text{L}^{-1/2}\text{T}^{-1}\) in fact). Further, since the free space permittivity is 1, in vacuo there is no distinction between \(\textbf{D}\) and \(\textbf{E}\), and either can substitute for the other. So the conversion between CGS esu and SI for \(\textbf{D}\) is the same as for \(\textbf{E}\)? No! In SI, we recognize \(\textbf{D}\) and \(\textbf{E}\) as being physically different quantities, and \(\textbf{D}\) is expressed in coulombs per square metre. Awful mental gymnastics are needed to find the conversion, but I'll give the answer here:

    \[ 1 \ \text{CGS esu of} \ \textbf{D} = \frac{10^5}{4 \pi c} \ \text{C m}^{-2}.\]

    Once again, please don't blame me – I'm just the messenger! And be warned – it is going to get worse – much worse

    Potential Difference

    If the work required to move a charge of 1 esu from one point to another is 1 erg, the potential difference between the points is 1 esu of potential difference, or 1 statvolt.

    It is often said that an esu of potential difference is 300 volts, but this is just an approximation. The exact conversion is

    \[1 \ \text{statvolt} = 10^{-8} c \ \text{V}.\]


    If the potential difference across the plate of a capacitor is one statvolt when the capacitor holds a charge of one statcoulomb, the capacitance of the capacitor is one centimetre. (No – that's not a misprint.)

    \[ 1 \ \text{cm} = 10^9 c^{-2} \text{F}.\]

    Here is a sample of some formulas for use with CGS esu.

    Potential at a distance \(r\) from a point charge \(Q\) in vacuo = \(Q/r\).

    Field at a distance \(r\) in vacuo from an infinite line charge of \(\lambda \ \text{esu/cm} = 2 \lambda /r\).

    Field in vacuo above an infinite charged plate bearing a surface charge density of \(\sigma \ \text{esu/cm}^2 = 2 \pi \sigma\).

    An electric dipole moment \(\textbf{p}\) is, as in SI, the maximum torque experienced by the dipole in unit electric field. A debye is \(10^{-18}\) esu of dipole moment. The field at a distance \(r\) in vacuo along the axis of a dipole is \(2p/r\).

    Gauss's theorem: The total normal outward flux through a closed surface is 4\(\pi\) time the enclosed charge.

    Capacitance of a plane parallel capacitor = \(\frac{kA}{4 \pi d}\).

    Capacitance of an isolated sphere of radius \(a\) in vacuo = \(a\). Example: What is the capacitance of a sphere of radius 1 cm? Answer: 1 cm. Easy, eh?

    Energy per unit volume or an electric field \(= E^2/(8 \pi)\).

    One more example before leaving esu. You will recall that, if a polarizable material is placed in an electrostatic field, the field \(\textbf{D}\) in the material is greater than \(\epsilon_0 \textbf{E}\) by the polarization \(\textbf{P}\) of the material. That is, \(\textbf{D}= \boldsymbol{\epsilon} \textbf{E} + \textbf{P}\). The equivalent formula for use with CGS esu is

    \[\textbf{D}=\textbf{E} + 4\pi \textbf{P}\]

    And since \(\textbf{P}= \chi_e \textbf{E}\) and \(\textbf{D} = k\textbf{E}\), it follows that

    \[k= 1 + 4 \pi \chi_e.\]

    At this stage you may want a conversion factor between esu and SI for all quantities. I'll supply one a little later, but I want to describe emu first, and then we can construct a table given conversions between all three systems.