What voltage must be applied to a 13.00-nanofarad capacitor to store 0.30 millicoulombs of charge?
In this example, we’re looking for a relationship between three quantities: charge, capacitance, and voltage. Voltage is what we want to solve for. These quantities are related through an equation for capacitance that capacitance is equal to charge divided by potential difference, voltage.
Since we want to solve not for capacitance but for voltage, we cross multiply and find that voltage is equal to charge over capacitance. Now it’s a matter of correctly plugging in the values of our charge 𝑄 and capacitance 𝐶. We’ll write that charge 𝑄 as 0.30 times 10 to the negative third coulombs. And the capacitance is 13.00 times 10 to the negative ninth farads.
When we calculate this fraction, to two significant figures, we find a result of 23 kilovolts. That’s the voltage that must be applied to this capacitor to store this amount 𝑄 of charge.