Help with Gaussian Units

[danmel413]

Homework Statement

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.

Homework 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)

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!

 

[Charles Link]

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.

 

[Charles Link]

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.

 

[Charles Link]

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.

 

[Charles Link]

It appears the OP has not returned to find out how to convert these units. This one was an interesting exercise that actually had a very simple answer. Most often, the conversion of electrical units from cgs to mks involves voltages. The conversion of capacitance per unit length isn't needed nearly as often (this is the first time I ever worked through it), but it has a simple and interesting solution.