Solutions

1)

From the vector-calculus identity $\nabla\cdot (f {\bf V}) = \nabla f\cdot {\bf V} + f\nabla \cdot {\bf V}$, we can write

So we can write

$$\nabla\cdot (vp\nabla u) = v\nabla\cdot (p\nabla u) + p\nabla v\cdot \nabla u~.$$ (1)

$$v{\cal L} u = \nabla\cdot (vp\nabla u)-p\nabla v\cdot \nabla u +vqu~,$$ (2)

and similarly for $u$ and $v$ switched. Using Eq. (2),

$$\int_{\cal V} d\tau ~(v {\cal L}u -u {\cal L}v) = \int_{\cal V} d\tau ~\nabla\cdot [ p(v\nabla u - u\nabla v ) ]~.$$ (3)

Using Stoke's theorem, Eq. (3) can be rewritten as the surface integral over the boundary: $\oint_{\cal S} d{\bf a}\cdot (v\nabla u - u \nabla v) p$.

2)

Using the superposition principle, the total electric potential can be expressed as

$$V(d) = \int {1\over 4\pi \epsilon_0} {dq\over r}~,$$ (4)

where $dq = \sigma Rd\phi dz$ in the cylindrical coordinate with $0<\phi < 2\pi$ and $-h< z<0$, $\sigma = {Q\over 2\pi Rh}$ is the surface charge density, and $r= \sqrt{ (d-z)^2 +R^2}$ is the distance to the charge source. If you explicitly spell out the integral, it is given as

$$V(d) = {1\over 4\pi \epsilon_0}\int_0^{2\pi} d\phi\int_{-h}^0 dz{ \sigma R \over \sqrt{ (d-z)^2 +R^2}}$$

$$= {Q\over 4\pi \epsilon_0 h } \ln \left[ { \sqrt{ d^2+R^2} -d \over \sqrt{(d+h)^2 +R^2} -(d+h)} \right]~,$$ (5)

using the integral formula $\int {dx\over \sqrt{ x^2+a^2} }= \ln ( x+\sqrt{x^2+a^2} )$.

The electric field is in the direction of the axis of the cylinder from the azimuthal symmetry, so $E_z = -\partial_d V(d)$, which gives

$$E_z = {Q\over 4\pi \epsilon_0 h } \left( {1\over \sqrt{d^2+R^2} } -{1\over \sqrt{(d+h)^2+R^2}} \right)~.$$ (6)

3)

Boundary conditions: $V(0,y,z) = V(x,0,z)= V(x,a,z)=0$ and $V(b,y,z) = V_0(y)$.

(a) Within the pipe, $\nabla ^2 V = 0$. Here one simplification we can make is to drop the $z$ dependence since the system is translationally invariant in $z$. So the equation reduces to $(\partial_x^2 +\partial_y^2) V = 0$.

Using the separation of variables technique in Cartesian coordinates (because the system is rectangular),

$$V = \sum_l v_l$$

where $v_l = X(x) Y(y)$. Then

$$\partial_x^2 X(x) =- \partial_y^2 Y(y) = l^2~,$$

since the function is expected to be periodic in $y$.

The solution is obtained as

$$Y = A_l \sin(ly) +B_l \cos (ly)$$

and

$$X = C_l \cosh (lx) +D_l \sinh( lx)$$

. From the boundary conditions $V(0,y,z) = V(x,0,z)= V(x,a,z)=0$, we must choose $B_l = C_l = 0$ and $l = n\pi/a$ where $n$ is an integer. So the form is expected to be

$$V = \sum_{n=1}^\infty c_n \sin ({n\pi y\over a} \sinh ({n\pi x\over a})~,$$

where $c_n$ is some coefficient (different from $C_l$ above).

We can determine the coefficients $c_n$ from the remaining boundary condition at $x=b$,

$$V_0(y) = \sum_{n=1}^\infty c_n \sin ({n\pi y\over a} \sinh ({n\pi b\over a})~,$$

which leads to

$$c_n = { \int _0^a dy~ V_0(y) \sin( {n\pi y\over a} ) \over {a\over 2} \sinh ( {n\pi b/over a}) }~.$$

(b) You simply replace $V_0(y) = V_0$, which gives

$$V(x,y) = {4V_0 \over \pi } \sum_{n \rm odd} { \sinh (n\pi x/a) \sin (n\pi y/a) \over n \sinh (n\pi b/a) }$$

4)

(A) Electric displacement: You can use the Gauss Law of displacement,

$$\oint {\bf D} \cdot d{\bf a} = Q_{\rm enc}~,$$ (7)

where $Q_{\rm enc}$ is the ``free'' charge enclosed by the Gaussian surface. Since Eq. (7) holds for both inside and outside the sphere, and $Q_{\rm enc}$ is the embedded charge $q$, you obtain ${\bf D}(r) = {q\over 4\pi r^2} \hat{r}$ for all $r$.

(B) Electric field: Since $E = D/\epsilon$ for a linear dielectric,

$${\bf E}(r) = {q\over 4\pi \epsilon_1 r^2} \hat{r}~~( r<R)$$ (8)

$$= {q\over 4\pi \epsilon_2 r^2} \hat{r}~~( r>R)~.$$ (9)

(C) Electric potential: The potential can be obtained from integrating the electric field from infinity as follows

$$V(r) = -\int_\infty ^r E(r) ~dr$$ (10)

$$= {q\over 4\pi \epsilon_2 r}~~(r>R)$$ (11)

$$= {q\over 4\pi \epsilon_2 R}+ {q\over 4\pi \epsilon_1 }\left( {1\over r} - {1\over R}\right)~~(r<R)~.$$ (12)

(D) Bound surface charge: It is the bound surface charge that makes the electric field discontinuous upon crossing the surface of the sphere at $r=R$. In other words, $\sigma_b = \epsilon_0 [ E(R+\delta) -E(R-\delta) ]$ for $\delta \rightarrow 0$. The result from (B) above can be used to calculate the following:

$$\sigma_b = {q\epsilon_0 \over 4\pi R^2} \left( {1\over \epsilon_2 }-{1\over \epsilon_1} \right)~.$$ (13)