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8.1: Charge and Energy

As far as particles go, we're familiar with the momentum and energy of charges. With fields, we have a similar situation. The fields determine what the particles do, and together with the energy/momentum of the particles themselves, the energy/momentum of the fields form a set of conservation laws

Conservation of Charge

We know that charge is conserved. You can't create or destroy charge by itself - you must compensate by destroying or creating some other charge. It's captured in the equation of continuity, which relates charge to the divergence of current. In laymen's terms, it says "if a charge was here, it must still be here or it must flow away." You can relate the amount of charge that's missing from a given region of space to the amount of charge that flows away from that element of space. This is a statement of conservation of local charge, not just global charge, which is a much stronger statement.

Say we have some charge distribution $\rho(\vec{r}, t)$ over a region S, and we have some current defined over the boundary $\vec{J}(\vec{r}, t)$. The relation between $\rho$ and $\vec{J}$ gives the equation of continuity. The change of charge within the entire volume must be compensated for by an amount of charge flowing in/out of the system:

$$\dv{Q}{t} = \dv{}{t} \int_{V} \rho \dd \tau = - \oint \vec{J} \cdot \dd a$$

We get a negative sign because by convention the normal to the surface points outward. We can take the time integral inside, and use Gauss's theorem (divergence theorem) on the right-side

$$\int_V \pdv{\rho}{t} \dd \tau = - \int_V (\div \vec{J}) \dd \tau$$

Since we chose an arbitrary volume, it must be the case that the integrands are equal, over all space.

$$\pdv{\rho}{t} = - \div \vec{J} \tagl{8.1}$$

Let's check that this is consistent with our Maxwell equations. Ampere law: $$\curl \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \pdv{\vec{E}}{t}$$ Take divergence of both sides $$\div( \curl \vec{B} ) = \mu_0 \div( \vec{J} ) + \mu_0 \epsilon_0 \div \left( \pdv{\vec{E}}{t} \right) \\ 0 = \mu_0 \div( \vec{J} ) + \mu_0 \epsilon_0 \pdv{}{t} \left( \div \vec{E} \right) \\ 0 = \mu_0 \div( \vec{J} ) + \mu_0 \epsilon_0 \pdv{\rho}{t} \frac{1}{\epsilon_0} \\ \rightarrow \pdv{\rho}{t} + \div \vec{J} = 0$$ So we're back to the continuity equation!

Conservation of Energy and Poynting's Theorem

The energy conservation we'll find is a combination of the energy stored in / done on the charges and the energy stored in the field. The new concept is the Poynting vector, describing the spatial flow of energy.

The Maxwell's equations tell us how to get from distributions of charges and currents to fields $\vec{E}$ and $\vec{B}$. The given distributions must satisfy the continuity relation, of course. Let's take a look at the amount of work that's done on the source charges, and use the result to connect to "where" the energy is, in some volume.

$$\vec{F} = q [ \vec{E} + \vec{v} \cross \vec{B} ]$$ $$\dd W = \vec{F} \cdot \dd \vec{l} = q ( \vec{E} + \vec{v} \cross \vec{B} ) \cdot \vec{v} \dd t$$ We want to generalize to a charge distribution $\dd q = \rho \dd \tau$ $$\dv{W}{t} = \int_V \vec{E} \cdot [ (\rho \dd \tau) \vec{v} ] = \int_V (\vec{E} \cdot \vec{J}) \dd \tau$$ Can we get an expression for $\dd W$ in terms of the fields only? Let's try and eliminate $\vec{F}$ in terms of $\vec{E}$ and $\vec{B}$ using Maxwell's equations. $$\vec{J} = \frac{\curl \vec{B}}{\mu_0} - \epsilon_0 \pdv{\vec{E}}{t}$$ $$\dv{W}{t} = \int_V \left( \frac{\vec{E} \cdot (\curl \vec{B}) }{\mu_0} - \epsilon_0 \vec{E} \cdot \pdv{\vec{E}}{t} \right) \dd \tau$$

Using vector identity $$\curl (\vec{A} \cross \vec{B}) = \vec{B} \cdot (\curl \vec{A}) - \vec{A} \cdot (\curl \vec{B})$$ $$\dv{W}{t} = \int_V \left( \frac{1}{\mu_0} \left( \vec{B} \cdot (\curl \vec{E}) - \div (\vec{E} \cross \vec{B}) \right) - \epsilon_0 \vec{E} \cdot \pdv{\vec{E}}{t} \right) \dd \tau \\ = \int_V \left( - \frac{\vec{B}}{\mu_0} \cdot \pdv{\vec{B}}{t} - \div \vec{S} - \epsilon_0 \vec{E} \cdot \pdv{\vec{E}}{t} \right) \dd \tau$$ where $\vec{S}$ is the so-called Poynting vector defined as $$\vec{S} = \frac{\vec{E} \cross \vec{B}}{\mu_0}$$ Let's take a look at the quantity $\dv{}{t} (\vec{E} ^2)$ $$\dv{}{t} \left( \vec{E}^2 \right) = \vec{E} \cdot \pdv{\vec{E}}{t} + \pdv{\vec{E}}{t} \cdot \vec E \quad \rightarrow \quad \vec E \cdot \pdv{\vec E}{t} = \frac{1}{2} \dv{}{t} \left( \vec E ^2 \right)$$ $$\dv{W}{t} = \int_V \left( - \frac{1}{2 \mu_0} \pdv{}{t} \left( \vec{B} ^2 \right) - \frac{\epsilon_0}{2} \pdv{}{t} \left( \vec E ^2 \right) - \div \vec{S} \right) \dd \tau \\ = - \dv{}{t} \left[ \int_V \frac{\vec B ^2}{2 \mu_0} \dd \tau + \int_V \frac{\epsilon_0 \vec E ^2}{2} \dd \tau \right] - \int_V \div \vec{S} \dd \tau$$ The integrands in the middle are just the energy densities of the electric and magnetic fields $$\frac{\vec B ^2}{2 \mu_0} = u_m \qquad \frac{\epsilon_0 \vec E ^2}{2} = u_e \qquad u_{em} \equiv u_e + u_m = u$$ $$\dv{W}{t} = - \dv{}{t} \int_V u_{em} \dd \tau - \oint _S \vec{S} \cdot \dd \vec a \tagl{8.2}$$ So what does this tell us? The left hand side reads "the change in energy in a region V". The first part of the r.h.s is the change in the local energy stored in the E and B fields, and the second part is an acknowledgment that our region is part of a larger space, and energy may be flowing in/out of the region. The flow of energy is therefore given by the Poynting vector, so we call $\eqref{8.1}$ "Poynting's Theorem"

In a charge-free region: $$0 = - \dv{}{t} \int_V u_{em} \dd \tau - \int_V ( \div \vec S ) \cdot \dd \tau \\ 0 = - \int_V \pdv{u_{em}}{t} \dd \tau - \int_V ( \div \vec S) \dd \tau$$ Since this is true for any arbitrary V, we have $$\rightarrow \pdv{u_{em}}{t} + \div \vec{S} = 0 \quad \text{(in charge-free region)}$$

Let's look at an example of the application of this statement of conservation of energy.

Problem 8.2

Consider the charging capacitor from problem 7.34. (a) Find the electric and magnetic fields in the gap, as functions of the distance s from the axis at time t (assume the charge is zero at t = 0). (b) Find the energy density $u_{em}$ and the Poynting vector $\vec S$ in the gap. Note especially the direction of $\vec S$. Check that $\eqref{8.2}$ is satisfied. (c) Determine the total energy in the gap, as a function of time. Calculate the total power flowing into the gap, by integrating the Poynting vector over the appropriate surface. Check that the power input is equal to the rate of increase of energy in the gap.

(a) From the ch. 7 problems, we know $$\vec B = \frac{\mu_0 I}{2 \pi a^2} s \hat{\phi}$$ Treating the gap as a parallel-plate capacitor, the electric field in the gap is $$\vec E = \frac{\sigma}{\epsilon_0} \hat{z} = \frac{I t}{\epsilon_0 \pi a^2} \hat{z}$$

(b) $$u_{em} = \frac{\vec B ^2}{2 \mu_0} + \frac{\epsilon_0 ^2 \vec E ^2}{2} \\ = \frac{1}{2} \epsilon_0 \frac{ I^2 t^2}{\epsilon_0 \pi^2 a^4} + \frac{\mu_0 ^2 I^2}{2 \mu_0 4 \pi ^2 a^4} s^2 \\ u_{em} = \frac{1}{2} \frac{I^2}{\pi ^2 a^4} \left[ \frac{t^2}{\epsilon_0} + \frac{\mu_0 s^2}{4} \right]$$ $$\vec S = \frac{1}{\mu_0} (\vec E \cross \vec B)$$ First, let's check what the direction of $\vec S$ is. We know that $\vec B$ is azimuthal and $\vec E$ is axial, so $\vec S$ must point radially inward. Working through the expression, we get $$\vec S = - \frac{1}{\epsilon_0} \frac{I^2 s t}{2 \pi^2 a^4} \hat{s}$$ Checking Poynting's theorem... $$\pdv{u_{em}}{t} = \frac{I^2 t}{\pi^2 a^4 \epsilon_0}$$ $$\div \vec S = \frac{1}{s} \pdv{}{s} \left( s S_s \right) = \frac{I^2 t}{\pi^2 a^4 \epsilon_0} \quad \checkmark$$ (c) The power flowing into the gap: we add a negative sign, since conventionally the normal direction is outwards, and the poynting vector points inwards. $$- \frac{1}{\mu_0} \int (\vec E \cross \vec B) \cdot \dd \vec a = + \frac{1}{\epsilon_0} \frac{I^2 s t}{2 \pi^2 a^4} \cross 2 \pi a w \\ = \frac{1}{\epsilon_0} \frac{I^2 t w}{\pi a^2}$$ The rate of increase of energy in the gap would be the time integral of the volume integral of the energy density $u_{em}$ in the gap. Again, the area of the cylinder is $\pi a^2 w$ $$\dv{}{t} \left[ \int u_{em} \dd \tau \right] = \dv{}{t} \frac{I^2}{2 \pi^2 a^4} \frac{t^2}{\epsilon_0} \cdot \pi a^2 w \\ = \frac{I^2 t}{\epsilon_0 \pi a^2 } = - \oint \vec S \cdot \dd \vec a \quad \checkmark$$