Calculate the voltage applied to a 2.00-microfarad capacitor when it holds 3.10 microcoulombs of charge.
In this example, we have three quantities that we want to connect: charge, capacitance, and voltage, what we want to solve for. And indeed, there is a relationship that brings all these three together. The capacitance of a capacitor is equal to the charge on the capacitor divided by the voltage across the plates.
We don’t want to solve for capacitance. But we do want to solve for voltage, which we can do by rearranging this equation. Cross multiplying, we find that the voltage across the capacitor is equal to the charge on the capacitor divided by its capacitance 𝐶. And these two quantities, charge and capacitance, are given to us in the problem statement. We write out the charge 𝑄 as 3.10 times 10 to the negative sixth coulombs and the capacitance 𝐶 as 2.00 times 10 to the negative sixth farad.
To three significant figures, this fraction is equal to 1.55 volts. That’s the voltage applied to this capacitor.