Video: Calculating the Capacitance Required to Store a Particular Charge

What capacitance is needed to store 3.00 ๐œ‡C of charge at a voltage of 120 V?

01:30

Video Transcript

What capacitance is needed to store 3.00 microcoulombs of charge at a voltage of 120 volts?

We can call the charge value of 3.00 microcoulombs ๐‘„. And letโ€™s name the voltage value, 120 volts, ๐‘‰. We want to solve for the capacitance of the system thatโ€™s needed to store this much charge ๐‘„ at this much voltage ๐‘‰. Weโ€™ll call that capacitance ๐ถ. Letโ€™s start off by recalling the mathematical definition of capacitance.

Capacitance is the capability of a system to store some amount of charge ๐‘„ at some voltage level ๐‘‰. The more charge there is in a system, the more capacitance that system has to contain all that charge. On the other hand, as potential difference decreases in a system, capacitance goes up in order to account for the small potential difference separating the charges in the system. The capacitance of our system is ๐‘„ over ๐‘‰ or 3.00 times 10 to the negative sixth coulombs divided by 120 volts.

When we calculate this fraction, we find that, to three significant figures, the systemโ€™s capacitance is 25.0 nanofarads. Thatโ€™s the capacitance needed to store this much charge at this much voltage.

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