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What is the energy stored in the 10.0μF capacitor of a heart defibrillator charged to 9.00×10³ V

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Final answer:

The energy stored in a 10.0-µF capacitor charged to 9.00 × 10³ V is 405 J. The amount of stored charge in the capacitor is 90 mC.

Step-by-step explanation:

Energy Stored in a Capacitor

The energy stored in a capacitor can be calculated using the formula:

E = ½ × C × V²

where E is the energy in joules (J), C is the capacitance in farads (F), and V is the voltage in volts (V). In the case of a 10.0-µF capacitor charged to 9.00 × 10³ V, we substitute the values into the formula:

E = ½ × 10.0 × 10 µF × (9.00 × 10³ V)²

Therefore, the energy stored is:

E = ½ × 10.0 × 10 µF× (9.00 × 10³)² = ½ × 10.0 × 10µF × 81.0 × 10µF = 405 J

Stored Charge in a Capacitor

The charge Q stored in a capacitor is calculated using the formula:

Q = C × V

For the given values, the amount of stored charge is:

Q = 10.0 × 10⁻µF × 9.00 × 10³ V = 90 mC

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