Question

In open-heart surgery, a much smaller amount of energy will defibrillate the heart.

(a) What voltage is applied to the 8.00 μF capacitor of a heart defibrillator that stores 40.0 J of energy?

a) 500 V
b) 200 V
c) 100 V
d) 50 V
Find the amount of stored charge.

a) 5.00 μC
b) 20.0 μC
c) 2.50 μC
d) 10.0 μC

Answer 1

To find the voltage and stored charge for the capacitor, we use the formulas ½ CV2 for energy and Q = CV for charge. However, the calculated voltage does not match any given options, suggesting an error in the provided choices.

Step-by-step explanation:

To find the voltage applied to the 8.00 µF capacitor of a heart defibrillator that stores 40.0 J of energy, we use the formula for the energy stored in a capacitor, which is ½ CV2, where C is the capacitance and V is the voltage. Solving for V, we get:

V = √(2 × energy / capacitance)

V = √(2 × 40 J / 8.00 × 10−6 F)

V = √(80 J / 8.00 × 10−6 F)

V = √(107 J/F)

V = 3162.3 V (approximately)

However, since this answer is not one of the options provided, there might be a mistake in the question or the answer choices.

To find the amount of stored charge, we use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. From the previous calculation, we would use V to find Q. However, since V is not matching the provided options, we'll hypothetically take one of the provided voltages, say 500 V, to give an example calculation:

Q = 8.00 µF × 500 V

Q = 4000 µC or 4 mC

This charge is larger than any of the provided options for stored charge, indicating there might be an inconsistency in the given choices.