Example 17-9 A Defibrillator Capacitor
A defibrillator, like the one shown in the photograph that opens this chapter, is essentially a capacitor that is charged by
a high-voltage source and then delivers the stored energy to a patient’s heart. (a) How much charge does the 80.0-mF
capacitor in a certain defibrillator store when it is fully charged by applying 2.50 kV? (b) How much energy can this defibrillator deliver?
Set Up
We’re given the capacitance C = 80.0 mF (recall
1 mF = 1 microfarad = 1026 F) and the potential difference V = 2.50 kV (recall 1 kV = 1
kilovolt = 103 V) between the capacitor plates.
We’ll use Equation 17-14 to determine the magnitude q of the charge on each capacitor plate,
and Equation 17-17 to find the potential energy
Uelectric stored in the charged capacitor. If we
assume that no energy is lost in the process of
being transferred to the patient, this is equal to
the energy that the defibrillator delivers.
Solve
(a) Substitute the given values of C and V into
Equation 17-14 to solve for the charge q.
Charge, voltage, and capacitance for a capacitor:
(17-14)
q = CV
Electric potential energy stored in a capacitor:
Uelectric =
q2
1
1
qV = CV 2 =
2
2
2C
(17-17)
Charge on the capacitor plates:
q = CV = (80.0 mF)(2.50 kV)
= (80.0 * 1026 F)(2.50 * 103 V)
= 0.200 F # V = 0.200 C
(Recall that 1 F = 1 C>V, so 1 F # V = 1 C.)
(b) Since the values of C and V are given,
let’s use the second of the three relationships
in Equation 17-17 to find the stored electric
potential energy.
Energy stored in the capacitor:
1
CV 2
2
1
= 180.0 * 10-6 F2 12.50 * 103 V2 2
2
= 2.50 * 102 F # V2
= 2.50 * 102 J
Uelectric =
(Recall that 1 V = 1 J>C and 1 F = 1 C>V, so 1 F # V 2 =
1 (C>V)V 2 = 1 C # V = 1 C # (J>C) = 1 J.)
Reflect
The American Heart Association recommends that a defibrillator shock should deliver
between 40 and 360 J in order to be effective,
so our numerical result is in the recommended
range.
We can double-check our answers by using
the other two relationships in Equation 17-17.
Happily, by using the value of q that we
One alternative way to calculate the energy stored in the capacitor:
calculated in part (a) we get the same result for
Uelectric in part (b) as we found above.
Remember that in solving any problem,
you should always take advantage of alternative ways to find the answer in order to check
your results.
Another alternative way to calculate the energy stored in the capacitor:
1
qV
2
1
= (0.200 C)(2.50 * 103 V)
2
= 2.50 * 102 C # V
= 2.50 * 102 J
Uelectric =
Uelectric =
=
q2
2C
(0.200 C)2
2(80.0 * 10-6 F)
C2
= 2.50 * 102
F
= 2.50 * 102 J
(Recall that 1 V = 1 J>C and 1 F = 1 C>V. You should verify that
1 C 2 >F = 1 J.)