In open heart surgery a much smaller amount of energy will defibrillate the heart.
(a) What voltage is applied to the 9.00 µF capacitor of a heart defibrillator that stores 45.0 J of energy?
kV
(b) Find the amount of stored charge.
mC
energy=1/2 C v^2 , right?
VQ=energy
To determine the voltage applied to the capacitor of the heart defibrillator and the amount of stored charge, we'll use the formula that relates energy, capacitance, and voltage:
Energy (E) = (1/2) * capacitance (C) * voltage (V)^2
(a) To find the voltage applied to the 9.00 µF capacitor:
Step 1: Rearrange the formula to solve for voltage (V):
V = √(2 * E / C)
Step 2: Substitute the given values into the formula:
E = 45.0 J (energy stored)
C = 9.00 µF = 9.00 × 10^-6 F (capacitance)
Step 3: Calculate the voltage:
V = √((2 * 45.0 J) / (9.00 × 10^-6 F))
Using a calculator, the voltage is approximately 2.00 kV.
Therefore, the voltage applied to the capacitor is 2.00 kV.
(b) To find the amount of stored charge:
Step 1: Use the formula for charge (Q) stored in a capacitor:
Q = C * V
Step 2: Substitute the given values into the formula:
C = 9.00 µF = 9.00 × 10^-6 F (capacitance)
V = 2.00 kV = 2.00 × 10^3 V (voltage)
Step 3: Calculate the amount of stored charge:
Q = (9.00 × 10^-6 F) * (2.00 × 10^3 V)
Using a calculator, the amount of stored charge is approximately 18.0 mC.
Therefore, the amount of stored charge is 18.0 mC.