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 17.1.) 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 changed. (See Figure 17.1.) Capacitors are also used to supply energy for flash lamps on cameras. In all of these examples, the key idea is the same: a capacitor is a device that can store separated charge, and because separated charge involves electric potential energy, a capacitor can store energy and then release it quickly when needed.

Energy stored in a capacitor is electrical potential energy, and it is thus related to the charge [latex]Q[/latex] and voltage [latex]V[/latex] on the capacitor. We must be careful when applying the equation for electrical potential energy [latex]\Delta \text{PE}=q\Delta V[/latex] to a capacitor. Remember that [latex]\Delta \text{PE}[/latex] is the change in potential energy of a particular amount of charge [latex]q[/latex] going through a potential difference [latex]\Delta V[/latex]. For a capacitor, however, the potential difference is not instantly at its final value when charging begins. A capacitor starts uncharged, so at the beginning it has essentially zero voltage across its plates. As charge is added, the plates become more strongly charged, the electric field between them becomes stronger, and the voltage across the capacitor increases. In other words, the “hill” that later charges must climb gets steeper as the capacitor fills up.

One way to see this clearly is to imagine adding charge to the capacitor in many tiny pieces. When the capacitor has very little charge on it, its voltage is very small, so pushing a tiny additional amount of charge onto the plates requires very little work. Later, after much more charge has accumulated, the capacitor’s voltage is larger, meaning that any additional charge you add must be moved against a larger potential difference. Thus the work required to add charge is not constant throughout the process—it increases as the capacitor charges up.

The first charge placed on a capacitor experiences a change in voltage [latex]\Delta V=0[/latex], since the capacitor has zero voltage when uncharged. The final charge placed on a capacitor experiences [latex]\Delta V=V[/latex], since the capacitor now has its full voltage [latex]V[/latex] on it. Because the voltage increases steadily from 0 to [latex]V[/latex] during charging, the average voltage during the process is [latex]V/2[/latex]. If we use the idea “energy stored equals (charge moved) × (average voltage),” then the energy stored in the capacitor, [latex]E_{\text{cap}}[/latex], is the total charge [latex]Q[/latex] multiplied by the average voltage [latex]V/2[/latex]. That is why the factor of 1/2 appears:

where [latex]Q[/latex] is the charge on a capacitor with a voltage [latex]V[/latex] applied. (Note that the energy is not [latex]QV[/latex], but [latex]QV/2[/latex].) This “one-half” is a reminder that the capacitor does not sit at voltage [latex]V[/latex] the whole time it is being charged; it rises from 0 to [latex]V[/latex]. Another way to interpret the result is to compare it to storing energy in other systems: for example, when you stretch a spring gradually from zero force to a final force, the energy stored is also not simply (final force)×(final displacement) but includes a factor of 1/2 because the force increases from 0 to its final value. The capacitor behaves similarly: the “electrical push” (voltage) increases as charge accumulates.

Charge and voltage are related to the capacitance [latex]C[/latex] of a capacitor by [latex]Q=\text{CV}[/latex], and so the expression for [latex]{E}_{\text{cap}}[/latex] can be written in forms that are often more convenient depending on what you know. If you know [latex]C[/latex] and [latex]V[/latex], substitute [latex]Q=\text{CV}[/latex] into [latex]E_{\text{cap}}=\frac{QV}{2}[/latex] to get [latex]E_{\text{cap}}=\frac{CV^{2}}{2}[/latex]. If instead you know [latex]Q[/latex] and [latex]C[/latex], you can eliminate [latex]V[/latex] using [latex]V=Q/C[/latex], giving [latex]E_{\text{cap}}=\frac{Q^{2}}{2C}[/latex]. These three expressions are completely equivalent and simply emphasize different measurable quantities:

where [latex]Q[/latex] is the charge and [latex]V[/latex] the voltage on a capacitor [latex]C[/latex]. The energy is in joules for a charge in coulombs, voltage in volts, and capacitance in farads. Each form can be useful: [latex]\frac{QV}{2}[/latex] highlights the idea of “charge moved through a potential difference,” [latex]\frac{CV^{2}}{2}[/latex] is especially convenient when a known voltage source charges a capacitor of known capacitance, and [latex]\frac{Q^{2}}{2C}[/latex] can be useful when you know how much charge has been stored but the voltage is not directly given.

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. To appreciate what the capacitor is doing here, remember that the heart’s muscle cells contract in response to coordinated electrical signals that travel through the heart tissue. During fibrillation, that coordination breaks down: electrical activity becomes disorganized, and different parts of the heart muscle can twitch out of sync rather than producing a strong, effective pump. A defibrillator attempts to stop that disorganized electrical activity by delivering a brief, high-energy pulse, essentially forcing many heart cells to depolarize at the same time. When the pulse ends, the hope is that the body’s natural pacemaker (and conduction system) can reestablish an organized rhythm.

From the physics point of view, the capacitor is crucial because it can store a substantial amount of energy and then release it very rapidly. A battery is excellent for providing moderate power over long times, but it is not designed to dump hundreds of joules in a fraction of a second. A charged capacitor, however, can discharge quickly through the defibrillator circuitry and the patient, delivering a controlled burst of energy. The device’s setting—such as “400 J”—refers to how much energy is stored in the capacitor (and then made available for the shock), which can be adjusted by changing the charging voltage or by using capacitors of particular capacitance. Because the stored energy depends on [latex]V^{2}[/latex] through [latex]E_{\text{cap}}=\frac{1}{2}CV^{2}[/latex], even modest changes in voltage can significantly change the stored energy.

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 17.2). 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. In practice, the shape of the pulse (the waveform) and the way the device matches the patient’s electrical resistance are important for delivering energy effectively and safely, but the underlying energy-storage mechanism is still the same: energy is stored electrically in a capacitor and then released quickly when needed.