Some Answers

Jackson (1.7)

Here we assume that the conducting wires are far apart, hence the charge is uniformly distributed over the surface of each wire. Hence the potential is given by

$$V(\mathbf{x}) = \frac{\lambda}{2\pi\epsilon_0} \log\left(\frac{r_+}{r_-} \right)$$ (1)

Thus potential of the two wires are $V_1=\frac{\lambda}{2\pi\epsilon_0} \log\left(\frac{a_1}{d-a_1} \right)$ and $V_2=\frac{\lambda}{2\pi\epsilon_0} \log\left(\frac{d-a_2}{a_2} \right)$. capacitance is

$$\frac{\lambda}{\vert v_1-v_2\vert} = \frac{2\pi\epsilon_0}{... ...\approx \frac{\pi\epsilon_0}{\vert\ln(\sqrt{a_1a_2}/d)\vert}$$ (2)

Jackson (1.10)

By integral formula (Eq 1.36 of Jackson) for potentials

$$\phi(\v{x}) = \frac{1}{4\pi}\oint\left(\frac{\partial\phi}{... ...artial n'} \left(\frac{1}{\vert\v{x}-\v{x'}\vert}\right) dS'$$ (3)

Since $\vert\v{x}-\v{x'}\vert=R$ and $\frac{\partial}{\partial n'} \left(\frac{1}{\vert\v{x}-\v{x'}\vert}\right)=-1/R^2$, the first integral vanishes by Gauss law and second integral is simply the average value of the potential on the surface of the sphere.

Jackson (1.12)

Hint: Use Green's second identity. Remember $\rho=-\epsilon_0\nabla^2\phi$ and $\sigma = +\epsilon_0\nabla\phi\cdot\v{\hat{n}'}$.

Jackson (2.2)

(a) The image charge of magnitude $-qa/y$ must be kept outside at a distance $a^2/y$. (d) All you need to do is add a constant to the result of part (a). The induced charge densities are same!

Jackson (2.3)

(a) Equipotential surfaces are shown in the figure.

Figure 1: Jackson 2.3 (a) Equipotential surfaces.(b) Charge density along x axis

(b) The required plots.

Jackson (2.4)

(a) The force on charge $q$ is given by

$$\v{F}_q = \frac{q^2\v{\hat{k}}}{4\pi\epsilon_0}\left[ \frac{d+R}{d^2} - \frac{dR}{(d^2-R^2)^2} \right]$$ (4)

Equating magnitude of the force to zero, we get three real solutions for $d/R=-0.61803,0.7549,1.61803$. Only one solution is outside the sphere.

Jackson (2.7)

The required Green's Function is

$$G(\v{x},\v{x'}) = \frac{1}{\vert\v{x}-\v{x'}\vert} - \frac{1}{\vert\v{x}-\v{x'}+2(\v{x'}\cdot\v{\hat{k}})\v{\hat{k}}\vert}$$ (5)

In cylindrical coordinates,

$$\frac{\partial G}{\partial n'} = -\frac{2z}{(r'^2+r^2+z^2-2rr'\cos(\phi-\phi'))^{3/2}}$$ (6)

Then, the potential at any point ($z>0$) is given by

$$\phi(\v{x}) = \frac{zV}{2\pi}\int_0^{2\pi}d\phi'\int_0^R r'dr' \frac{1}{(r'^2+r^2+z^2-2rr'\cos(\phi-\phi'))^{3/2}}$$ (7)