The electric potential energy stored in the capacitor of adefibrillator is 69 J,

Question

The electric potential energy stored in the capacitor of adefibrillator is 69 J, and the capacitanceis 115 µF. What is the potentialdifference across the capacitor plates?
____________________________V
my professor said to use these equations: Energy=1/2 Q V=1/2C V^2=Q^2/2C I simply don't understand
The electric potential energy stored in the capacitor of adefibrillator is 69 J, and the capacitanceis 115 µF. What is the potentialdifference across the capacitor plates?
____________________________V
my professor said to use these equations: Energy=1/2 Q V=1/2C V^2=Q^2/2C I simply don't understand

Explanation / Answer

W=(1/2)CV2 V=(2W/C) = (2*69J/115*10-6F) =1.095*103 V = 1095 V potential difference across the capacitor plates is1095V V=(2W/C) = (2*69J/115*10-6F) =1.095*103 V = 1095 V potential difference across the capacitor plates is1095V

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