Question

(a) What is the energy stored in the $10.0\textrm{ }\mu\textrm{F}$ capacitor of a heart defibrillator charged to $9.00 \times 10^3 \textrm{ V}$? (b) Find the amount of stored charge.

- $405 \textrm{ J}$
- $0.0900 \textrm{ C}$

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Video Transcript

This is College Physics Answers with Shaun Dychko. The energy stored in the heart defibrillator will be the capacitance times the voltage squared divided by two. And, that's going to be ten microfarads, which is ten times ten to the minus six farads times nine times ten to the three volts squared divided by two, which is 405 Joules. To find the amount of charge stored in the defibrillator, we'll say that the capacitance is defined basically as the amount of charge that can be accumulated divided by the voltage applied to make that charge get there. And, we'll solve this with

*Q*, by multiplying both sides by*V*. And then, switch the sides around. And, we get*Q*is*C**V*. So, that's ten microfarads times 9000 volts, which gives 0.0900 Coulombs, the charge.