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How to Convert Capacitance in cm to Farads for a Transmission Line Experiment?
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[QUOTE="Charles Link, post: 5503107, member: 583509"] 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. [/QUOTE]
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How to Convert Capacitance in cm to Farads for a Transmission Line Experiment?
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