Chapter 10: Potentials and Fields

Check that the potentials of a point charge moving at constant velocity (Eqs. 10.49 and 10.50) satisfy the Lorenz gauge condition (Eq. 10.12).

For the configuration in Ex. 10.1, consider a rectangular box of length $l$, width $\mathbf{w}$, and height $\mathbf{h}$, situated a distanced $\mathbf{d}$above the $\mathbf{yz}$plane (Fig. 10.2).

Figure 10.2

(a) Find the energy in the box at time${\mathbf{t}}_1=\mathbf{d}/\mathbf{c}$, and at${\mathbf{t}}_2=(\mathbf{d}+\mathbf{h})/\mathbf{c}$.

(b) Find the Poynting vector, and determine the energy per unit time flowing into the box during the interval${\mathbf{t}}_1<\mathbf{t}<{\mathbf{t}}_2$.

(c) Integrate the result in (b) from ${\mathbf{t}}_1$to ${\mathbf{t}}_2$, and confirm that the increase in energy (part (a)) equals the net influx.

A particle of charge $q_1$is at rest at the origin. A second particle, of charge$q_2$ , moves along the axis at constant velocity $v$.

(a) Find the force $F_{12}\left(t\right)$ of${\mathit{q}}_{\mathbf{1}}$ on $q_2$, at time$t$ . (When$q_2$ is at $z=vt$).

(b) Find the force $F_{21}\left(t\right)$of$q_2$ on$q_1$ , at time $t$. Does Newton's third law hold, in this case?

(c) Calculate the linear momentum$p\left(t\right)$ in the electromagnetic fields, at time$t$ . (Don't bother with any terms that are constant in time, since you won't need them in part (d)). [Answer:$({\mu }_0{q}_1{q}_2/4\pi t)$ ]

(d) Show that the sum of the forces is equal to minus the rate of change of the momentum in the fields, and interpret this result physically.

Develop the potential formulation for electrodynamics with magnetic charge (Eq. 7.44).

Find the (Lorenz gauge) potentials and fields of a time-dependent ideal electric dipole $\mathbf{p}\left(\mathbf{t}\right)$ at the origin. (It is stationary, but its magnitude and/or direction are changing with time.) Don't bother with the contact term. [Answer:

$\begin{array}{l}\mathbf{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_0}\frac{\widehat{r}}{{\mathbf{r}}^2}\mathbf{\cdot }\mathbf{[}\mathbf{p}\mathbf{+}\mathbf{(}\mathbf{r}\mathbf{/}\mathbf{c}\mathbf{)}\dot{\mathbf{p}}\mathbf{]}\\ \mathbf{A}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{{\mathbf{\mu }}_0}{\mathbf{4}\mathbf{\pi }}\left[\frac{}{}\right]\\ \mathbf{E}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{-}\frac{{\mathbf{\mu }}_0}{\mathbf{4}\mathbf{\pi }}\left\{\frac{\ddot{P}-\widehat{r}(\widehat{r}\cdot \ddot{p})}{}+c^2\frac{[p+(r/c)\dot{p}]-3\widehat{r}(\widehat{r}\cdot [p+(r/c)\dot{p}])}{r^3}\right\}\\ \mathbf{B}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{-}\frac{{\mathbf{\mu }}_0}{\mathbf{4}\mathbf{\pi }}\left\{\frac{\widehat{r}\times [\dot{p}+(r/c)\ddot{p}]}{r^2}\right\}\end{array}$

Where all the derivatives of $\mathbf{p}$ are evaluated at the retarded time.]

A piece of wire bent into a loop, as shown in Fig. 10.5, carries a current that increases linearly with time:

$\mathit{I}\left(t\right)\mathbf{=}\mathit{k}\mathit{t}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(-\infty <t<\infty \right)$

Calculate the retarded vector potential A at the center. Find the electric field at the center. Why does this (neutral) wire produce an electric field? (Why can’t you determine the magnetic field from this expression for A?)

Suppose$\mathit{J}\left(r\right)$ is constant in time, so (Prob. 7.60 ) $\mathit{p}\left(r,t\right)\mathbf{=}\mathit{p}\left(r,0\right)\mathbf{+}\mathit{p}\left(r,0\right)\mathit{t}$. Show that

$\mathit{E}\left(r,t\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\int }\frac{\mathbf{p}\left(r\text{'},t\right)}{{\mathbf{r}}^{\mathbf{2}}}\hat{\mathbf{r}}\mathit{d}\mathcal{b}\mathbf{\text{'}}$

that is, Coulomb’s law holds, with the charge density evaluated at the non-retarded time.

Suppose the current density changes slowly enough that we can (to good approximation) ignore all higher derivatives in the Taylor expansion

$\mathit{J}\left(t_r\right)\mathbf{=}\mathit{J}\left(t\right)\mathbf{+}\left(t_r-t\right)\mathit{J}\left(t\right)\mathbf{+}\mathbf{\dots }$

(for clarity, I suppress the r-dependence, which is not at issue). Show that a fortuitous cancellation in Eq. 10.38 yields

$\mathit{B}\left(r,t\right)\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\mathbf{\int }\frac{\mathbf{J}\left(r\text{'},t\right)\mathbf{\times }\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}\mathit{d}\mathcal{b}\mathbf{\text{'}}$.

That is: the Biot-Savart law holds, with J evaluated at the non-retarded time. This means that the quasistatic approximation is actually much better than we had any right to expect: the two errors involved (neglecting retardation and dropping the second term in Eq. 10.38 ) cancel one another, to first order.

A particle of chargeq moves in a circle of radius a at constant angular velocity $\mathit{\omega }$. (Assume that the circle lies in thexy plane, centered at the origin, and at time$\mathit{t}\mathbf{=}\mathbf{0}$ the charge is at role="math" localid="1653885001176" $\left(\mathit{a}\mathbf{,}\mathbf{0}\right)$, on the positive x axis.) Find the Liénard-Wiechert potentials for points on the z-axis.

Show that the scalar potential of a point charge moving with constant velocity (Eq. 10.49) can be written more simply as

$\mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{\mathbf{R}\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{v}}^{\mathbf{2}}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }}{{\mathbf{c}}^{\mathbf{2}}}}}}$ (10.51)

where$\mathit{R}\mathbf{\equiv }\mathit{r}\mathbf{-}\mathit{v}\mathit{t}$is the vector from the present (!) position of the particle to the field point r, and$\mathit{\theta }$is the angle between R and v (Fig. 10.9). Note that for nonrelativistic velocities $(v^2\ll c^2)$,

$\mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{\approx }\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{\mathbf{R}}$