Question

capacite 1: 7.3x10^-3 charge unit transfer
capacite 2: 0. 1 C/v

That means equating the two capacitances gives the conversion factor for converting between these two different units of capacitance. Follow this course to determine the number of Coulombs in one chargetransfer-unit. Show your work.




Now calculate how many fundamental units of charge transfer each time you add charges to the capacitor. Show your work.



If we assume it is electrons being transferred, does this number represent the number of electrons being transferred to the capacitor or removed from the capacitor? Explain.

I need help with these questions. please and thank you

Answer 1

To calculate the conversion factors:

We equate capacitance 1 to capacitance 2.

Therefore,

And,

Adding a single fundamental charge to the capacitor increases the charge by:

Then, for unit increase in the charge transfer unit, the number of fundamental charges that need to be added:

This is the number of electrons added or removed from the capacitor. The charge of an electron has the same magnitude as e but the opposite polarity. Adding electrons would imply that one of the plates of the capacitor accumulates more negative charge. But this is accompanied by an equal increase in positive charge in the other plate of the capacitor. This is exactly how a capacitor stores charge. The two plates are charge equally with opposite polarity. As for removal of electrons, the other plate will develop and equal negative charge. And the number n of electrons when added or removed, the charge in the capacitor increases by 1 charge-transfer unit x Volt. (Note that charge transfer unit is a unit of capacitance not charge, so it must be multiplied with the unit volt to give dimensions of charge).