# Homework Help: Help with Gaussian Units

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1. Jun 20, 2016

### danmel413

1. The problem statement, all variables and given/known data
Basically I have to convert from Gaussian Units in a Transmission Line Experiment. I don't get what the point of expressing Capacitance in cm is and I find this very confusing.

2. Relevant equations
The equation I've been given is C'=Q'/V=w/4πh, (1) where the prime denotes "per unit length" and w and h are the width of the line and the distance between the plates, equal to 2.6 and 2.7 cm respectively.

Conversion: In a poorly denoted tabled, the conversion is given as 1F = (c2/1E9) cm (2)

3. The attempt at a solution

So as far as I can tell it appears that it's a linear conversion by using the factor given in (2), which is pretty much c2/1E9 (units of cm/sec) which is about 8.94E11, and I multiply this by w/4πh, however I fail to see how this results in a Capacitance in Farads. Since the units of length cancel in (1), a) how is this a per unit length and b) how would this conversion not result in a speed instead of capacitance? Clearly I'm not factoring something in but I don't know what.

Thanks for any help!

2. Jun 20, 2016

The voltage from a single proton at a distance of 1 centimeter is $\vec{V}=(Q_{cgs}/r_{cgs}) \hat{V_{cgs}}$ in c.g.s. and $\vec{V}=(Q_{mks}/((4 \pi \epsilon_o)r_{mks}) \hat{V_{mks}}$ in mks so that $(4.8 E-10/1) \hat{V_{cgs}}=((1.602 E-19)(9.0 E+9)/.01) \hat{V_{mks}}$. Thereby $1 \hat{V_{cgs}}=300 \hat{V_{mks}}$. Capacitance C=Q/V. Capacitance per unit length will be $C_l=Q/(Vd)$. Writing $Q=C_l V d$ we can write the expression in mks units and cgs units for the same system. Now one more item: Voltage $\vec{V}=V_{cgs} \hat{V_{cgs}}=V_{mks} \hat{V_{mks}}$ so that $V_{cgs}=(1/300) V_{mks}$. Similarly distance $\vec{d}=d_{cgs} (1 cm)=d_{mks} (1 meter)$. Since $1meter=100cm$, $d_{cgs}=100 d_{mks}$. Let's consider a system that holds some charge Q: $C_l(cgs) V_{cgs} d_{cgs}=Q_{cgs}=((4.8E-10)/(1.602 E-19))Q_{mks}=(3.0E+9) (C_l(mks) V_{mks} d_{mks})$. Putting it all together (with the expressions for $d_{cgs}$ and $V_{cgs}$ ), I get $C_l(mks)=((1/9)E-9) C_l(cgs)$. So compute $C_l (cgs)$ with your formula and the mks capacitance per unit length $\vec{C_l}(mks)=((1/9)E-9) C_l(cgs)$ in farads/meter. A somewhat lengthy process, but quite systematic and hopefully my algebra/arithmetic is correct.

Last edited: Jun 20, 2016
3. Jun 21, 2016

Additional item: Looking over the above conversion factor, it does appear that the formula for capacitance per unit length in mks units is likely to read $C_l(mks)=(w/(4 \pi h))4\pi \epsilon_o=(w/h) \epsilon_o$ but it would take a little more work to show this by a process similar to the above calculations. (The factor $1/(4 \pi \epsilon_o)=9.0E+9$ is well known in the Coulomb's law equation.) The capacitance per unit length is dimensionless in the cgs system, but in the mks system, this $4 \pi \epsilon_o$ factor would also supply the necessary units...editing...In fact, writing for a point charge $V=Q/r$ (c.g.s.) and $V=Q/(4 \pi \epsilon_o r)$ (mks), it is apparent that any expression involving capacitance $C=Q/V$ in mks units is going to have a $4 \pi \epsilon_o$ in it that is not present in the cgs expression.

Last edited: Jun 21, 2016
4. Jun 21, 2016

Upon further inspection, the capacitance per unit length $C/L=Q/(VL)=w/(4 \pi h)$ in cgs units is actually a simple formula that is easy to derive (using Gauss's law) for two parallel conducting plates of length L and width w and a distance h apart. In mks units, the capacitance per unit length for the same parallel plates is given $C/L=(w/h) \epsilon_o$. The above calculations (posts #2 and #3) are confirmed by simply deriving the formula for capacitance per unit length that was provided.

Last edited: Jun 21, 2016
5. Jun 24, 2016