Solutions

1)

(a) Close to $r=0$, the potential is

(b) (i) Inside and outside of the void, $\nabla ^2 \phi =0$. (ii) Boundary conditions: $\phi (R+\delta) =\phi(R-\delta)$ where $\delta \rightarrow 0$, and from $\nabla\cdot {\bf D} =0$, $\epsilon_0\partial_r \phi(R-\delta) =\epsilon \partial_r \phi(R+\delta)$.

(c) From (a),

$$\phi(r) ={1\over 4\pi \epsilon_0} {p\cos \theta \over r^2}~.$$ (21)

$$\phi = {1\over 4\pi \epsilon_0} {p\cos \theta \over r^2} +\sum _{l=0}^\infty A_l r^l P_l(\cos\theta )~,(r<R)$$ (22)

$$= \sum_{l=0}^\infty {B_l\over r^{l+1}}P_l(\cos\theta )~. (r>R)$$ (23)

The boundary conditions determine the coefficients $A_l,B_l$:

$$A_1 = {2p\over 4\pi\epsilon_0R^3}{ \epsilon_0-\epsilon \over 2\epsilon +\epsilon_0}~,$$ (24)

$$B_1 = {p\over 4\pi \epsilon_0}{ 2\epsilon_0+\epsilon \over 2\epsilon+\epsilon_0 }~,$$ (25)

and all the other coefficients are zero.

2)

(a) In the $(\omega , {\bf k})$ domain, the Maxwell's equations are

$${\bf k\cdot E} = 0~,$$ (26)

$$i{\bf k\times E} = i\omega {\bf B}~,$$ (27)

$${\bf k\cdot B} = 0~,$$ (28)

$$i{\bf k\times B} = \mu {\bf J} -i\omega \epsilon {\bf E}~.$$ (29)

Here ${\bf J}=0$. By eliminating ${\bf B}$, you can obtain the wave equation $(k^2-\mu\epsilon \omega ^2) {\bf E} =0$, and the same for ${\bf B}$ after eliminating ${\bf E}$.

(b) Since ${\bf J} = \sigma {\bf E} \ne 0$, the wave equation becomes $(k^2-\mu\epsilon \omega ^2-i\omega \mu\sigma ){\bf E} =0$.

(c) From (b), the dispersion is obtained as $k = \sqrt{ \omega ^2 \mu \epsilon (\omega) +i\omega \mu\sigma }$. If you expand this to the first order in $\chi$, you obtain

$$k \approx \omega \sqrt{ \mu \epsilon_0 +i \mu\sigma/\omega }... ...\omega) \over 2( \epsilon_0\mu +i \mu\sigma/\omega)} \right]~.$$ (30)

Since $\chi(\omega )$ is the factor that exhibits the anomalous dispersion, the correction term to the absorption coefficient will still show the anomalous dispersion in such a conductor.

3)

(a) $B=\mu_0 nI_s$ inside the solenoid. So the emf is

$${\cal E} = -d\Phi/dt = \pi a^2 \mu_0 n{dI_s\over dt} = I_rR~,$$ (31)

which gives

$$I_r = {\pi a^2 \mu_0 n\over R}{dI_s\over dt}~.$$

(b) The electric field at the solenoid surface satisfies $\oint dl E = 2\pi aE ={\cal E}$, and the magnetic fiels at the axis of the ring is

$$B ={ \mu_0I_r b^2 \over 2 (b^2+z^2)^{3/2} }~.$$

Therefore, the Poynting vector is

$${\bf S} = {1\over \mu_0} {\bf E\times B}~.$$ (32)

The total power emanating from the solenoid is $P = \int {\bf S} \cdot d{\bf a} = I_r^2Rb^2 \int_0^\infty dz/(b^2+z^2)^{3/2} = i_r^2 R$.

4)

(a)

$${\bf E} = {q\over 4\pi\epsilon_0 [ x-x(t_r)]^2}\left[ {c+v(t_r)\over c-v(t_r) }\right]\hat{x}~,$$ (33)

where $t_r = t-[x-x(t_r)]/c$ and $v(t) =dx(t)/dt$. The magnetic field is zero.

(b) $P = \mu_0 q^2a^2/6\pi c$ where $a= d^2x(t)/dt^2$. Time-averaged, $\bar{P} = \omega ^4 \mu_0 q^2x_0^2/12\pi c$.

(c) The radiation damping force is $F_{rad} = (\mu_0 q^2/6\pi c) da/dt$. If you take the time integral of $F_{rad} v(t)$, then you obtain $\bar{P}$. In other words, the energy loss of the particle due to the damping is the radiated power.