97.5k views
5 votes
What is the energy (in J) stored in the 14.0 µF capacitor of a heart defibrillator charged to 6.90 ✕ 103 V?

1 Answer

3 votes

Answer:

The energy stored in the capacitor is 333.3 J

Step-by-step explanation:

Energy stored in a capacitor = CV²/2

C = capacitance = 14 × 10⁻⁶F

V = 6.90 × 10³V

Energy = (14 × 10⁻⁶)(6.90 × 10³)²/2 = 333.27 = 333.3 J

Hope this Helps!!!

User Kelvin Zhao
by
4.6k points