Find the charge stored when 5.50 V is applied to an 8.00 pF capacitor.
OpenStax College Physics, Chapter 19, Problem 47 (Problems & Exercises)
This is College Physics Answers with Shaun Dychko. Capacitance is defined as the amount of charge that can be accumulated in the capacitor per voltage. And so, we can multiply both sides by voltage to solve for Q. So, Q is the capacitance times voltage, and eight picofarads is eight times ten to the minus 12 farads, and we multiply that by 5.50 volts to get 4.40 times ten to the minus 11 Coulombs of charge accumulated.