In open heart surgery a much smaller amount of energy will defibrillate the heart.
(a) What voltage is applied to the 5.00 µF capacitor of a heart defibrillator that stores 35.0 J of energy?
kV
(b) Find the amount of stored charge.
mC
To find the voltage applied to the 5.00 µF capacitor, we can use the formula for the energy stored in a capacitor:
E = ½ * C * V^2
where E is the energy, C is the capacitance, and V is the voltage.
(a) Solving for V, we can rearrange the formula:
V^2 = (2 * E) / C
V = sqrt((2 * E) / C)
Substituting the given values:
E = 35.0 J
C = 5.00 µF = 5.00 x 10^(-6) F
V = sqrt((2 * 35.0 J) / (5.00 x 10^(-6) F))
Calculating this, we get:
V ≈ 469.3 V
So, the voltage applied to the 5.00 µF capacitor is approximately 469.3 volts.
(b) To find the amount of stored charge, we can use the formula:
Q = C * V
where Q is the charge, C is the capacitance, and V is the voltage.
Substituting the given values:
C = 5.00 µF = 5.00 x 10^(-6) F
V = 469.3 V
Q = (5.00 x 10^(-6) F) * (469.3 V)
Calculating this, we get:
Q ≈ 2.35 x 10^(-3) C
So, the amount of stored charge is approximately 2.35 milliCoulombs.