Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Charging a capacitor takes work, and the capacitor can give this work back: a camera flash, a defibrillator and the back-up supply of a memory chip all rely on it. Where is the energy kept in the meantime? We shall answer this question in two steps: first we compute the work needed to assemble a set of charges, then we rewrite the result so that the energy appears to be spread through the field. The final answer is

W = \frac{1}{2} \int_{v} ε E^{2} d v = \frac{1}{2} \int_{v} ε E \cdot E d v = \frac{1}{2} \int_{v} D \cdot E d v [J], (3.4.24)

the integral extending over all space where the field exists. In a homogeneous medium $ε$ can be taken outside the integral, $W = \frac{ε}{2} \int_{v} ∣ E ∣^{2} d v$ . Rather than accept Eq. (3.4.24) as a definition, let us derive it from the work needed to assemble a charge distribution.

The work of assembling charges

Assembling charges one at a time, slowly, from infinity. The first charge costs nothing. Each later charge is pushed against the field of those already in place, and the work is the charge times the potential they produce at its final position (V_{kj}: potential of Q_j at the position of Q_k). A negative Q_3 is pulled in by Q_1 and Q_2, so its contribution W_3 is negative. — **Figure 3.4.22.** Assembling charges one at a time, slowly, from infinity. The first charge costs nothing. Each later charge is pushed against the field of those already in place, and the work is the charge times the potential they produce at its final position ( $V_{k j}$ : potential of $Q_{j}$ at the position of $Q_{k}$ ). A negative $Q_{3}$ is pulled in by $Q_{1}$ and $Q_{2}$ , so its contribution $W_{3}$ is negative.

Start with an empty space and bring point charges in from infinity, one at a time, so slowly that no kinetic energy is involved (Figure 3.4.22). Bringing the first charge $Q_{1}$ costs nothing, since there is no field yet. Bringing $Q_{2}$ to its place costs $W_{2} = Q_{2} V_{21}$ , where $V_{21} = Q_{1} / (4 π ε_{0} R_{12})$ is the potential of $Q_{1}$ at the position of $Q_{2}$ (Section 3.3: the work done against the field is charge times potential difference, and $V = 0$ at infinity). Bringing $Q_{3}$ costs $W_{3} = Q_{3} (V_{31} + V_{32})$ , and so on. For $N$ charges

W = k = 2 \sum N j < k \sum Q_{k} V_{k j}, V_{k j} = \frac{Q _{j}}{4 π ε _{0} R _{k j}} .

Since $Q_{k} V_{k j} = Q_{k} Q_{j} / (4 π ε_{0} R_{k j}) = Q_{j} V_{j k}$ , each pair appears once and the sum is symmetric. Writing it as half of the sum over all ordered pairs $j \neq = k$ (each pair is then counted twice, hence the $\frac{1}{2}$ ),

W = \frac{1}{2} k = 1 \sum N Q_{k} j \neq = k \sum V_{k j} = \frac{1}{2} k = 1 \sum N Q_{k} V_{k},

where $V_{k}$ is the potential at the position of $Q_{k}$ due to all the other charges. For a continuous distribution the sum becomes an integral:

Energy in terms of charge and potential

W = \frac{1}{2} k = 1 \sum N Q_{k} V_{k} ⟶ W = \frac{1}{2} \int_{v} ρ_{v} V d v . (3.4.25)

(In the continuous form $V$ is the potential of the whole distribution. The contribution of an infinitesimal element to the potential at its own position is negligible, which is why the restriction $j \neq = k$ disappears. For true point charges the continuous form would include their infinite self-energies. See Exercise 3.4.6.)

Energy in terms of the field

Now we move the energy from the charges to the field. The plan: replace $ρ_{v}$ by $\nabla \cdot D$ , integrate by parts with a vector identity, and show that the surface term vanishes at infinity. Using $\nabla \cdot (V D) = V \nabla \cdot D + D \cdot \nabla V$ :

W = \frac{1}{2} \int_{v} V = ρ_{v} \nabla \cdot D d v = \frac{1}{2} \int_{v} \nabla \cdot (V D) d v - \frac{1}{2} \int_{v} D \cdot \nabla V d v .

The volume $v$ may be taken as a sphere of radius $R$ so large that it contains all the charges. By the divergence theorem the first integral becomes $\frac{1}{2} \oint_{S} V D \cdot d S$ . Far from the charges $V$ falls off at least as $1/ R$ and $D$ at least as $1/ R^{2}$ , while the area grows as $R^{2}$ . The surface integral falls off at least as $1/ R$ and vanishes as $R \to \infty$ . In the second integral $- \nabla V = E$ . Hence, integrating over all space,

Energy in terms of the field

W w = \frac{1}{2} \int_{v} D \cdot E d v = \frac{1}{2} \int_{v} ε E^{2} d v [J], = \frac{1}{2} D \cdot E = \frac{1}{2} ε E^{2} = \frac{D ^{2}}{2 ε} [J/ m^{3}] . (3.4.26)

The two expressions (3.4.25) and (3.4.26) give the same total energy but describe it differently: $\frac{1}{2} \int ρ_{v} V d v$ counts the work of assembling the charges, while $\frac{1}{2} \int D \cdot E d v$ locates the same energy in the field, with the energy density $w$ at each point. The field picture is the one that survives in electrodynamics, where fields carry energy away from their sources. Exercise 3.4.6 compares the two for a uniformly charged sphere.

In plain words

Wherever there is an electric field there is stored energy, $\frac{1}{2} ε E^{2}$ joules in every cubic metre. Strong fields store a lot (the energy grows with the square of $E$ ), so most of the energy sits where the field is strongest.

Where the energy of a charged conducting sphere (radius R, in air) is stored. The energy density w=\tfrac12\varepsilon_0E^2 falls off as 1/r^4 (orange), yet the volume of a shell grows as r^2: the energy stored between R and r is the fraction 1-R/r of the total (blue). Half of the energy lies within r=2R, the other half in the whole of space beyond (Exercise 3.4.5). — **Figure 3.4.23.** Where the energy of a charged conducting sphere (radius $R$ , in air) is stored. The energy density $w = \frac{1}{2} ε_{0} E^{2}$ falls off as $1/ r^{4}$ (orange), yet the volume of a shell grows as $r^{2}$ : the energy stored between $R$ and $r$ is the fraction $1 - R / r$ of the total (blue). Half of the energy lies within $r = 2 R$ , the other half in the whole of space beyond (Exercise 3.4.5).

Energy of a charged capacitor

For a system of conductors, each conductor is an equipotential carrying the surface charge $Q_{k}$ , and Eq. (3.4.25) becomes $W = \frac{1}{2} \sum_{k} Q_{k} V_{k}$ . For two conductors carrying $+ Q$ and $- Q$ at the potentials $V_{1}$ and $V_{2}$ ,

W V = \frac{1}{2} Q V_{1} - \frac{1}{2} Q V_{2} = \frac{1}{2} Q V, = V_{1} - V_{2} . (3.4.27)

With the capacitance $C = Q / V$ defined in Section 3.4.10, this is $W = \frac{1}{2} Q V = \frac{1}{2} C V^{2} = Q^{2} / (2 C)$ .

Recipe · Computing the stored energy in two ways

From the circuit quantities: find $Q$ and $V$ (or $C$ ) and use $W = \frac{1}{2} Q V = \frac{1}{2} C V^{2} = Q^{2} / (2 C)$ . For a charge distribution use $W = \frac{1}{2} \int ρ_{v} V d v$ .
From the field: find $E$ (and $D$ ) everywhere, write $w = \frac{1}{2} ε E^{2}$ with the local $ε$ , choose the volume element of the symmetry ( $S d x$ , $2 π ρ L d ρ$ , $4 π r^{2} d r$ ) and integrate region by region.
Compare: the two results must agree. If they do not, look for a forgotten region (the field outside a charged sphere), a wrong $ε$ in one layer, or a missing factor $\frac{1}{2}$ .

Common mistake · Forgetting the

\frac{1}{2}

The stored energy is $\frac{1}{2} Q V$ , not $Q V$ . The source that charges a capacitor to the voltage $V$ does deliver $Q V$ , but half of it is dissipated in the resistance of the charging circuit (or radiated), whatever that resistance is (compare Exercise 3.4.8). The factor $\frac{1}{2}$ comes from the charge being brought in against a potential that grows from $0$ to $V$ .

Quick check

A capacitor connected to a source has its voltage doubled. By what factor does the stored energy change? And the energy density between the plates?

Both grow by $4$ : $W = \frac{1}{2} C V^{2}$ and $w = \frac{1}{2} ε E^{2}$ with $E \propto V$ .

Application 3.4.2 · Electrostatic energy storage

The principle of a defibrillator. A capacitor is charged slowly from a high-voltage supply (switch S_1) and then discharged through the patient's chest in a few milliseconds (switch S_2). Right: the exponential discharge current. — **Figure 3.4.24.** The principle of a defibrillator. A capacitor is charged slowly from a high-voltage supply (switch S $_{1}$ ) and then discharged through the patient's chest in a few milliseconds (switch S $_{2}$ ). Right: the exponential discharge current.

In a capacitor the field fills only the dielectric between the plates, so by Eq. (3.4.26) the stored energy is $W = \frac{1}{2} ε E^{2} \times (volume of the dielectric)$ , and in terms of circuit quantities it is $W = \frac{1}{2} C V^{2} = \frac{1}{2} Q V$ by Eq. (3.4.27). There are two quite different routes to a large stored energy.

High field, ordinary capacitance: the film capacitor

A defibrillator (Figure 3.4.24) charges a capacitor slowly and discharges it into the patient's chest in milliseconds. External shocks are of the order of $100$ – $360$ J. With voltages in the kilovolt range, $W = \frac{1}{2} C V^{2}$ keeps the capacitance at tens of microfarads: $200$ J at $4$ kV requires $C = 2 W / V^{2} = 25$ μF. Polypropylene film operated at $E \approx 200$ MV/m stores

\frac{1}{2} ε E^{2} = \frac{1}{2} \times 2.2 \times 8.842 \times 1 0^{- 12} \times (2 \times 1 0^{8})^{2} \approx 0.39 J/c m^{3},

so the 200 J need about half a litre of film, wound into a roll (Worked example 3.4.2). With the chest resistance $R \approx 50 Ω$ , the time constant is $R C = 1.25$ ms and the peak current $V / R = 80$ A: the energy is delivered in a few milliseconds.

Low voltage, enormous capacitance: the supercapacitor

In an electrochemical double-layer capacitor (supercapacitor) the "plates" are a porous carbon electrode and a layer of ions of the electrolyte that gathers about $1$ nm from its surface. Porous carbon has a surface area of the order of $1 0^{3}$ m $^{2}$ per gram, so $C = εS / d$ reaches the order of $1 0^{2}$ F per gram of electrode, and cells of $3000$ F are common. The electrolyte decomposes above roughly $2.7$ V, which limits the energy to $W = \frac{1}{2} \times 3000 \times 2. 7^{2} \approx 11$ kJ $\approx 3$ Wh per cell. This is a few watt-hours per kilogram, far below the $150$ – $250$ Wh/kg of lithium-ion batteries, but a supercapacitor survives $1 0^{5}$ – $1 0^{6}$ charge cycles and delivers kilowatt pulses.

The two devices illustrate one rule: the energy density $\frac{1}{2} ε E^{2}$ is set by the field the dielectric can bear. The supercapacitor stores some fifty times more energy than the defibrillator capacitor, but at $2.7$ V it could drive only $2.7/50 = 54$ mA through the chest. For a defibrillator, what counts is the voltage.

$W = \frac{1}{2} C V^{2}$ , the energy density $\frac{1}{2} ε E^{2}$ is limited by $E_{bd}$

3.4.9 Electrostatic energy

The work of assembling charges

Energy in terms of the field

Energy of a charged capacitor

High field, ordinary capacitance: the film capacitor

Low voltage, enormous capacitance: the supercapacitor