College Physics for AP® Courses

# 19.7Energy Stored in Capacitors

College Physics for AP® Courses19.7 Energy Stored in Capacitors

### Learning Objectives

By the end of this section, you will be able to:

• List some uses of capacitors.
• Express in equation form the energy stored in a capacitor.
• Explain the function of a defibrillator.

The information presented in this section supports the following AP® learning objectives and science practices:

• 5.B.2.1 The student is able to calculate the expected behavior of a system using the object model (i.e., by ignoring changes in internal structure) to analyze a situation. Then, when the model fails, the student can justify the use of conservation of energy principles to calculate the change in internal energy due to changes in internal structure because the object is actually a system. (S.P. 1.4, 2.1)
• 5.B.3.1 The student is able to describe and make qualitative and/or quantitative predictions about everyday examples of systems with internal potential energy. (S.P. 2.2, 6.4, 7.2)
• 5.B.3.2 The student is able to make quantitative calculations of the internal potential energy of a system from a description or diagram of that system. (S.P. 1.4, 2.2)
• 5.B.3.3 The student is able to apply mathematical reasoning to create a description of the internal potential energy of a system from a description or diagram of the objects and interactions in that system. (S.P. 1.4, 2.2)

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 19.29.) 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 19.29.) Capacitors are also used to supply energy for flash lamps on cameras.

Figure 19.29 Energy stored in the large capacitor is used to preserve the memory of an electronic calculator when its batteries are charged. (credit: Kucharek, Wikimedia Commons)

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

$Ecap=QV2,Ecap=QV2, size 12{E rSub { size 8{"cap"} } =Q { {V} over {2} } } {}$
19.74

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:

$E cap = QV 2 = CV 2 2 = Q 2 2C , E cap = QV 2 = CV 2 2 = Q 2 2C , size 12{E rSub { size 8{"cap"} } = { { ital "QV"} over {2} } = { { ital "CV" rSup { size 8{2} } } over {2} } = { {Q rSup { size 8{2} } } over {2C} } } {}$
19.75

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.

### Energy Stored in Capacitors

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

$E cap = QV 2 = CV 2 2 = Q 2 2C , E cap = QV 2 = CV 2 2 = Q 2 2C , size 12{E rSub { size 8{"cap"} } = { { ital "QV"} over {2} } = { { ital "CV" rSup { size 8{2} } } over {2} } = { {Q rSup { size 8{2} } } over {2C} } } {}$
19.76

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. Energy stored in the capacitor is internal potential energy.

### Making Connections: Point Charges and Capacitors

Recall that we were able to calculate the stored potential energy of a configuration of point charges, and how the energy changed when the configuration changed in Applying the Science Practices: Work and Potential Energy in Point Charges. Since the charges in a capacitor are, ultimately, all point charges, we can do the same with capacitors. However, we write it down in terms of the macroscopic quantities of (total) charge, voltage, and capacitance; hence Equation (19.76).

For example, consider a parallel plate capacitor with a variable distance between the plates connected to a battery (fixed voltage). When you move the plates closer together, the voltage still doesn’t change. However, this increases the capacitance, and hence the internal energy stored in this system (the capacitor) increases. It turns out that the increase in capacitance for a fixed voltage results in an increased charge. The work you did moving the plates closer together ultimately went into moving more electrons from the positive plate to the negative plate.

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 19.30). 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.

Figure 19.30 Automated external defibrillators are found in many public places. These portable units provide verbal instructions for use in the important first few minutes for a person suffering a cardiac attack. (credit: Owain Davies, Wikimedia Commons)

### Example 19.11

#### Capacitance in a Heart Defibrillator

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

$Ecap=CV22.Ecap=CV22. size 12{E rSub { size 8{"cap"} } = { { ital "CV" rSup { size 8{2} } } over {2} } } {}$
19.77

#### 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 "." {} } } {}
19.78

#### Discussion

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

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