Most of us have seen dramatizations in which medical personnel use a defibrillator to pass an electric current through a patient’s heart to get it to beat normally. (Review Figure 1.) Often realistic in detail, the person applying the shock directs another person to “make it 400 joules this time.” The energy delivered by the defibrillator is stored in a capacitor and can be adjusted to fit the situation. SI units of joules are often employed. Less dramatic is the use of capacitors in microelectronics, such as certain handheld calculators, to supply energy when batteries are charged. (See Figure 1.) Capacitors are also used to supply energy for flash lamps on cameras.

Energy stored in a capacitor is electrical potential energy, and it is thus related to the charge QQ size 12{Q} {} and voltage VV size 12{V} {} on the capacitor. We must be careful when applying the equation for electrical potential energy ΔPE=qΔVΔPE=qΔV size 12{?"PE"=q?V} {} to a capacitor. Remember that ΔPEΔPE size 12{?"PE"} {} is the potential energy of a charge *qq size 12{q} {}* going through a voltage ΔVΔV size 12{?V} {}. But the capacitor starts with zero voltage and gradually comes up to its full voltage as it is charged. The first charge placed on a capacitor experiences a change in voltage ΔV=0ΔV=0 size 12{?V=0} {}, since the capacitor has zero voltage when uncharged. The final charge placed on a capacitor experiences ΔV=VΔV=V size 12{?V=V} {}, since the capacitor now has its full voltage VV size 12{V} {} on it. The average voltage on the capacitor during the charging process is V/2V/2 size 12{V/2} {}, and so the average voltage experienced by the full charge *qq size 12{q} {}* is V/2V/2 size 12{V/2} {}. Thus the energy stored in a capacitor, EcapEcap size 12{E rSub { size 8{"cap"} } } {}, is

Ecap=QV2,Ecap=QV2, size 12{E rSub { size 8{"cap"} } =Q { {V} over {2} } } {}

(1)where QQ size 12{Q} {} is the charge on a capacitor with a voltage VV size 12{V} {} applied. (Note that the energy is not QVQV size 12{ ital "QV"} {}, but QV/2QV/2 size 12{ ital "QV"/2} {}.) Charge and voltage are related to the capacitance
C
C
of a capacitor by Q=CVQ=CV size 12{Q= ital "CV"} {}, and so the expression for EcapEcap size 12{E rSub { size 8{"cap"} } } {} can be algebraically manipulated into three equivalent expressions:

E
cap
=
QV
2
=
CV
2
2
=
Q
2
2C
,
E
cap
=
QV
2
=
CV
2
2
=
Q
2
2C
,
size 12{E rSub { size 8{"cap"} } = { { ital "QV"} over {2} } = { { ital "CV" rSup { size 8{2} } } over {2} } = { {Q rSup { size 8{2} } } over {2C} } } {}

(2)where QQ size 12{Q} {} is the charge and VV size 12{V} {} the voltage on a capacitor CC size 12{C} {}. The energy is in joules for a charge in coulombs, voltage in volts, and capacitance in farads.

The energy stored in a capacitor can be expressed in three ways:

E
cap
=
QV
2
=
CV
2
2
=
Q
2
2C
,
E
cap
=
QV
2
=
CV
2
2
=
Q
2
2C
,
size 12{E rSub { size 8{"cap"} } = { { ital "QV"} over {2} } = { { ital "CV" rSup { size 8{2} } } over {2} } = { {Q rSup { size 8{2} } } over {2C} } } {}

(3)where QQ size 12{Q} {} is the charge, VV size 12{V} {} is the voltage, and CC size 12{C} {} is the capacitance of the capacitor. The energy is in joules for a charge in coulombs, voltage in volts, and capacitance in farads.

In a defibrillator, the delivery of a large charge in a short burst to a set of paddles across a person’s chest can be a lifesaver. The person’s heart attack might have arisen from the onset of fast, irregular beating of the heart—cardiac or ventricular fibrillation. The application of a large shock of electrical energy can terminate the arrhythmia and allow the body’s pacemaker to resume normal patterns. Today it is common for ambulances to carry a defibrillator, which also uses an electrocardiogram to analyze the patient’s heartbeat pattern. Automated external defibrillators (AED) are found in many public places (Figure 2). These are designed to be used by lay persons. The device automatically diagnoses the patient’s heart condition and then applies the shock with appropriate energy and waveform. CPR is recommended in many cases before use of an AED.

A heart defibrillator delivers
4.00
×
10
2
J
4.00
×
10
2
J
of energy by discharging a capacitor initially at
1.00
×
10
4
V
1.00
×
10
4
V
. What is its capacitance?

*Strategy*

We are given EcapEcap size 12{E rSub { size 8{"cap"} } } {} and VV size 12{V} {}, and we are asked to find the capacitance CC size 12{C} {}. Of the three expressions in the equation for EcapEcap size 12{E rSub { size 8{"cap"} } } {}, the most convenient relationship is

Ecap=CV22.Ecap=CV22. size 12{E rSub { size 8{"cap"} } = { { ital "CV" rSup { size 8{2} } } over {2} } } {}

(4)
*Solution*

Solving this expression for CC size 12{C} {} and entering the given values yields

C
=
2
E
cap
V
2
=
2
(
4
.
00
×
10
2
J
)
(
1
.
00
×
10
4
V
)
2
=
8
.
00
×
10
–
6
F
=
8
.
00 µF
.
C
=
2
E
cap
V
2
=
2
(
4
.
00
×
10
2
J
)
(
1
.
00
×
10
4
V
)
2
=
8
.
00
×
10
–
6
F
=
8
.
00 µF
.
alignl { stack {
size 12{C= { {2E rSub { size 8{"cap"} } } over {V rSup { size 8{2} } } } = { {"800"" J"} over { \( 1 "." "00"´"10" rSup { size 8{4} } " V" \) rSup { size 8{2} } } } =8 "." "00"´"10" rSup { size 8{-6} } " F"} {} #
"=8" "." "00 "mF "." {}
} } {}

(5)
*Discussion*

This is a fairly large, but manageable, capacitance at
1.00
×
10
4
V
1.00
×
10
4
V
.

Comments:"This introductory, algebra-based, two-semester college physics book is grounded with real-world examples, illustrations, and explanations to help students grasp key, fundamental physics concepts. […]"