## Energy Stored in Capacitors

The ability to store charge in a capacitor is only one aspect of its usefulness. The other aspect is its ability to store energy. The two aspects are related because it takes work to charge a capacitor. Charging really means that we create a charge imbalance between the plates by moving excess electrons (negative charges) onto one plate against their mutual repulsion, and simultaneously removing electrons from the other plate against the attractive force of the positively charged atomic nuclei that remain there.

Working against these forces is analogous to lifting a weight against gravity – except that gravity stays constant, whereas electric forces change depending on the amount of charge that has already been moved.

Read this text, which explores the relationship between stored charge and stored energy in a capacitor.

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 19.23.) 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 19.23.) Capacitors are also used to supply energy for flash lamps on cameras.

Figure 19.23 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 $Q$ and voltage $V$ on the capacitor. We must be careful when applying the equation for electrical potential energy $\Delta PE=q\Delta V$ to a capacitor. Remember that $\Delta PE$ is the potential energy of a charge $q$ going through a voltage $\Delta V$.

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 $\Delta V=0$, since the capacitor has zero voltage when uncharged. The final charge placed on a capacitor experiences $\Delta V=V$, since the capacitor now has its full voltage $V$ on it. The average voltage on the capacitor during the charging process is $V$, and so the average voltage experienced by the full charge $q$ is $V/2$. Thus the energy stored in a capacitor, $E_{cap}$, is

$E_{cap}=\frac{QV}{2}$ [equation 19.74]

where $Q$ is the charge on a capacitor with a voltage $V$ applied. (Note that the energy is not $QV$, but $QV/2$.) Charge and voltage are related to the capacitance $C$ of a capacitor by $Q=CV$, and so the expression for $E_{cap}$ can be algebraically manipulated into three equivalent expressions:

$E_{cap}=\frac{QV}{2}=\frac{CV^{2}}{2}=\frac{Q^{2}}{2C}$ [equation 19.75]

where $Q$ is the charge and $V$ the voltage on a capacitor $C$. 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:

$E_{cap}=\frac{QV}{2}=\frac{CV^{2}}{2}=\frac{Q^{2}}{2C}$ [equation 19.76]

where $Q$ is the charge, $V$ is the voltage, and $C$ 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 19.24). 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 19.24 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 $4.00\times 10^{2}\: J$ of energy by discharging a capacitor initially at $1.00\times 10^{4}\: V$. What is its capacitance?

#### Strategy

We are given $E_{cap}$ and $V$, and we are asked to find the capacitance $C$. Of the three expressions in the equation for $E_{cap}$, the most convenient relationship is

$E_{cap}=\frac{CV^{2}}{2}$ [equation 19.77]

#### Solution

Solving this expression for $C$ and entering the given values yields

$C=\frac{2E_{cap}}{V^{2}}=\frac{2(4.00\times 10^{2}J)}{(1.00\times10^{4}V)^{2}}=8.00\times 10^{-6}F$ [equation 19.78]

$=8.00\: \mu F$

#### Discussion

This is a fairly large, but manageable, capacitance at $1.00\times10^{4}$.

Source: Rice University, https://openstax.org/books/college-physics/pages/19-7-energy-stored-in-capacitors