3rd of September, 2008

1. $\stackrel{\to }{a}\perp \stackrel{\to }{b}$, so $\stackrel{\to }{a}\cdot \stackrel{\to }{b}=0$ and $\stackrel{\to }{v}\cdot \stackrel{\to }{a}=\alpha$. Multiplying the first equation by $\stackrel{\to }{a}$ from the left (using the cross product), we get $\stackrel{\to }{a}×\left(\stackrel{\to }{v}×\stackrel{\to }{a}\right)=\stackrel{\to }{a}×\stackrel{\to }{b}⇔{a}^{2}\stackrel{\to }{v}-\stackrel{\to }{a}\left(\stackrel{\to }{v}\cdot \stackrel{\to }{a}\right)=\stackrel{\to }{a}×\stackrel{\to }{b}$. Using that $\stackrel{\to }{v}\cdot \stackrel{\to }{a}=\alpha$,
$\stackrel{\to }{v}=\frac{\stackrel{\to }{a}×\stackrel{\to }{b}+\alpha \stackrel{\to }{a}}{{a}^{2}}$

1. $\nabla \cdot \left(\left(\stackrel{\to }{b}\cdot \stackrel{\to }{r}\right)\stackrel{\to }{a}\right)=\stackrel{\to }{b}\cdot \stackrel{\to }{a}\phantom{\rule{2em}{0ex}}\nabla ×\left(\left(\stackrel{\to }{b}\cdot \stackrel{\to }{r}\right)\stackrel{\to }{a}\right)=\stackrel{\to }{b}×\stackrel{\to }{a}$

2. $\nabla \cdot \stackrel{\to }{r}=3\phantom{\rule{2em}{0ex}}\nabla ×\stackrel{\to }{r}=0$

3. $\nabla \cdot \left(r\stackrel{\to }{r}\right)=4r\phantom{\rule{2em}{0ex}}\nabla ×\left(r\stackrel{\to }{r}\right)=0$

(Note: The curl of any radially symmetric vector field is 0.)

4. $\nabla \cdot \left(\stackrel{\to }{a}×\left(\stackrel{\to }{b}×\stackrel{\to }{r}\right)\right)=-2\stackrel{\to }{a}\cdot \stackrel{\to }{b}\phantom{\rule{2em}{0ex}}\nabla ×\left(\stackrel{\to }{a}×\left(\stackrel{\to }{b}×\stackrel{\to }{r}\right)\right)=\stackrel{\to }{a}×\stackrel{\to }{b}$

5. $\nabla \cdot \left(\left(\stackrel{\to }{a}\cdot \stackrel{\to }{r}\right)\left(\stackrel{\to }{b}×\stackrel{\to }{r}\right)\right)=\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right)\cdot \stackrel{\to }{r}\phantom{\rule{2em}{0ex}}\nabla ×\left(\left(\stackrel{\to }{a}\cdot \stackrel{\to }{r}\right)\left(\stackrel{\to }{b}×\stackrel{\to }{r}\right)\right)=3\stackrel{\to }{b}\left(\stackrel{\to }{a}\cdot \stackrel{\to }{r}\right)-\stackrel{\to }{r}\left(\stackrel{\to }{a}\cdot \stackrel{\to }{b}\right)$

6. $\nabla \cdot \left(\stackrel{\to }{r}×\left(\stackrel{\to }{a}×\stackrel{\to }{r}\right)\right)=-2\stackrel{\to }{a}\cdot \stackrel{\to }{r}\phantom{\rule{2em}{0ex}}\nabla \cdot \left(\stackrel{\to }{r}×\left(\stackrel{\to }{a}×\stackrel{\to }{r}\right)\right)=3\stackrel{\to }{r}×\stackrel{\to }{a}$

The curl of an electrostatic field is 0, so if $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are not parallel, then only the fields in points (b) and (c) may be electrostatic fields.

1. The charge density is
 $\begin{array}{cc}\begin{array}{rl}\rho & ={\epsilon }_{0}\nabla \cdot \stackrel{\to }{E}\\ & ={\epsilon }_{0}\nabla \cdot \left(A\frac{{e}^{-br}}{{r}^{3}}\stackrel{\to }{r}\right)\\ & ={\epsilon }_{0}A\left(\nabla {e}^{-br}\right)\frac{\stackrel{\to }{r}}{{r}^{3}}+{\epsilon }_{0}A{e}^{-br}\left(\nabla \cdot \frac{\stackrel{\to }{r}}{{r}^{3}}\right)\end{array}& \end{array}$

Let us calculate the two terms of this sum separately:

$\nabla {e}^{-br}=-b{e}^{-br}\stackrel{^}{r},$

where $\stackrel{^}{r}=\stackrel{\to }{r}∕r$, and

$\nabla \cdot \frac{\stackrel{\to }{r}}{{r}^{3}}=\nabla \cdot \frac{\stackrel{^}{r}}{{r}^{2}}=4\pi \delta \left(\stackrel{\to }{r}\right).$

Thus the complete charge density is

$\rho \left(\stackrel{\to }{r}\right)=4\pi {\epsilon }_{0}A\delta \left(\stackrel{\to }{r}\right)-{\epsilon }_{0}Ab\frac{{e}^{-br}}{{r}^{2}}.$

2. The total charge enclosed by a spherical surface $S$ of radius $r$ centred around the origin is
$Q\left(r\right)={\epsilon }_{0}{\oint }_{S}\stackrel{\to }{E}\cdot \stackrel{\to }{S}={\epsilon }_{0}A{e}^{-br}\frac{1}{{r}^{2}}\phantom{\rule{3.26212pt}{0ex}}4\pi {r}^{2}=4\pi {\epsilon }_{0}A{e}^{-br}.$

Thus as $r\to \infty$, the total charge enclosed by the surface goes to 0.

2. The work needed to add another layer of charge $dq$ to a sphere of radius $r$ and total charge $q$ is
$dW=\frac{1}{4\pi {\epsilon }_{0}}\frac{qdq}{r}$

The sphere has a homogeneous charge distribution, so $q=\frac{4\pi }{3}{r}^{3}\rho$ and $dq=4\pi \rho {r}^{2}dr$.

Thus the energy needed to assemble a charged sphere of radius $R$ and charge $Q$ is

${\int }_{0}^{R}dW=\frac{4\pi {\rho }^{2}{R}_{0}^{5}}{15{\epsilon }_{0}}=\frac{3}{20\pi {\epsilon }_{0}}\frac{{Q}^{2}}{R}.$

The electric field is

and the potential is

3. Let us find the gravitational potential energy $W\left(x\right)$ of the two half-spheres as a function of their separation $x$. The force keeping them together is simply $F=dW∕dx$.

Creating a gap by separating the halves is equivalent to removing (“mining out”) all the matter from the gap, and distributing it on the surface (i.e. it causes the same change $\Delta W$ in energy). Provided that the gap between the halves is very narrow compared to the planet’s radius ($x\ll R$), the change in the gravitational field is negligible.

Using Gauß’s law, it is easily shown that the gravitational potential below the surface of the spherical planet is

where $\gamma$ is the gravitational constant, $M$ is the mass of the planet, and $r$ is the distance from the centre.

When moving a piece of mass $dm$ up to the surface, the increase in energy is $\left(V\left(R\right)-V\left(r\right)\right)dm$. To get the total change in energy, we must integrate over the volume of the gap:

 (1)

The force keeping the halves together is

$F=\frac{dW\left(x\right)}{dx}=\frac{dW\left(x\right)}{dx}-\underset{0}{\underbrace{\frac{dW\left(0\right)}{dx}}}=\frac{d\Delta W}{dx}=\frac{3\gamma {M}^{2}}{16{R}^{2}}.$

(I.e. the same as the weight of an object of mass $\frac{3}{16}M$ on the surface of the planetoid.)

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