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 𝑄. And let’s name the voltage value, 120 volts, 𝑉. We want to solve for the capacitance of the system that’s needed to store this much charge 𝑄 at this much voltage 𝑉. We’ll call that capacitance 𝐶. Let’s start off by recalling the mathematical definition of capacitance.
Capacitance is the capability of a system to store some amount of charge 𝑄 at some voltage level 𝑉. 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 𝑄 over 𝑉 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 system’s capacitance is 25.0 nanofarads. That’s the capacitance needed to store this much charge at this much voltage.