Question

Calculate the voltage applied to a $2.00 \textrm{ }\mu\textrm{F}$ capacitor when it holds $3.10 \textrm{ }\mu\textrm{C}$ of charge.

$1.55 \textrm{ V}$

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Video Transcript

This is College Physics Answers with Shaun Dychko. Capacitance is the amount of charge accumulated divided by the voltage required to put that charge on the capacitor. And, we can solve for

*V*by multiplying both sides by*V*over*C*. And then, you get the voltage then as the charge divided by the capacitance. So, 3.10 microCoulombs is 3.10 times ten to the minus six Coulombs, and then divide by two microfarads, which is two times ten the minus six farads, and we end up with 1.55 volts must have been applied to this capacitor to accumulate that amount of charge.