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Up: PHY531-532 Classical Electrodynamics Links

PHY531 Problem Set 1. Due Sept 6, 2001.

N.B. All homework must be done in Gaussian units.

First a warm up exercise. Do not turn this in, but be sure to do it. Download my Fortran program surface.f from the class web site. My code calculates the integral of the normal component of the electric field of a unit point charge on the surface of an ellipsoid. Gauss's law tells us that we should get $4 \pi$ if the charge is inside the ellipsoid and zero if it is outside. Read every line in this code, and then run it for at least 5 cases where the charge is inside and 5 where it is outside the ellipsoid. To get you started, compile the code. On the linux machines you can do this with the command

g77 -O2 -o surface surface.f
Next you can run it and expect something like:
prompt%surface
  x y z of unit charge
1.0 2.0 3.0
  semiaxes lengths: a b c
3.0 4.0 5.0
  Nu Nv
100 100
 integral over ellipsoid/4 pi =   0.99981611
prompt%
which gives a result close to one as expected for a charge inside the ellipsoid. An example that puts the charge outside is:
prompt%surface
  x y z of unit charge
4.0 4.0 4.0
  semiaxes lengths: a b c
3.0 4.0 5.0
  Nu Nv
100 100
 integral over ellipsoid/4 pi =   3.15063198E-06
prompt%

  1. Jackson Problem 1.7, the result in Gaussian units is
    \begin{displaymath}
C \simeq \frac{1}{4 \ln \frac{d}{a} } \,.
\end{displaymath} (1)

  2. A filamentary circular current loop of radius $a$ is in the $x-y$ plane centered on the origin. The total current is $I$ and is in the counter clockwise direction when viewed from a position on the positive $z$ axis.
    a.
    Write an expression for ${\vec {J}} (x,y,z)$ in cartesian coordinates in terms of delta functions.
    b.
    Write an expression for ${\vec {J}} (\rho,\theta,z)$ in cylindrical coordinates in terms of delta functions.
    c.
    Write an expression for ${\vec {J}}(r,\theta,\phi)$ in spherical coordinates in terms of delta functions.
    d.
    Calculate explicitly
    \begin{displaymath}
{\vec {X}} \equiv \int d^3r {\vec {J}}({\vec {r}})
\end{displaymath} (2)

    using the expressions for ${\vec {J}}$ in each of the coordinate systems.

  3. Coulomb's law gives the electrostatic scalar potential to be
    \begin{displaymath}
\Phi({\vec {r}}) = \int d^3r' \frac{\rho({\vec {r'}})}{\vert{\vec {r}} - {\vec {r'}}\vert} \,.
\end{displaymath} (3)

    Let's assume that $\rho({\vec {r}})$ is spherically symmetric; that is it only depends on the distance $r$ from the origin and not on the direction, and can be written $\rho(r)$ where $r = \vert{\vec {r}} \vert$.
    a.
    Since you can pick your coordinate system, to calculate $\Phi$ at any particular ${\vec {r}}$ you can pick the coordinate system for ${\vec {r'}}$ with $z$ along ${\vec {r}}$. In that case,
    \begin{displaymath}
\vert{\vec {r}} - {\vec {r'}}\vert = \sqrt{r^2 + r'^2 - 2 r r' \cos \theta' }
\end{displaymath} (4)

    where $r' = \vert{\vec {r'}}\vert$. Show by direct integration that
    \begin{displaymath}
\int_{-1}^1 d\cos \theta' \int_0^{2\pi} d\phi'
[r^2 + r'^2 ...
...1/2}
=\frac{2\pi \left [ r+r'-\vert r-r'\vert \right ]}{r r'}
\end{displaymath} (5)

    and show that this can be written as
    \begin{displaymath}
\frac{2\pi \left [ r+r'-\vert r-r'\vert \right ]}{r r'} =\frac{4\pi}{r_>} \,.
\end{displaymath} (6)

    where $r_>$ is the larger of $r'$ and $r$.

    The potential is therefore

    \begin{displaymath}
\Phi({\vec {r}}) =
4\pi \int_0^\infty \frac{\rho(r')}{r_>} r...
...0^r \rho(r') r'^2 dr' +
4\pi \int_r^\infty \rho(r') r' dr' \,.
\end{displaymath} (7)

    b.
    You can also use Gauss's law to calculate the electric field $E_r(r)$, and by integrating obtain the electrostatic scalar potential. Show that the Gauss's law expression
    \begin{displaymath}
\Phi({\vec {r}}) = \int_r^\infty E_r(r') dr' \,.
\end{displaymath} (8)

    can be converted into the direct integration result above.

  4. In your math methods course you should have been (or will be) introduced to general curvilinear orthogonal coordinate systems. Let's apply the method to spheroidal coordinates. We use a metric tensor, but don't be afraid. The metric tensor here is your friend and you just have to plug into the standard formula, Eq 11, below. The metric tensor simplifies the ugly algebra you would otherwise get if you just ground through the transformation of derivatives using the chain rule.
    a.
    Calculate the metric tensor for the transformation to oblate spheroidal coordinates $(\xi,\gamma,\phi)$ defined by
    $\displaystyle x$ $\textstyle =$ $\displaystyle c \sqrt{1+\xi^2} \sin \gamma \cos \phi$  
    $\displaystyle y$ $\textstyle =$ $\displaystyle c \sqrt{1+\xi^2} \sin \gamma \sin \phi$  
    $\displaystyle z$ $\textstyle =$ $\displaystyle c~\xi~ \cos \gamma$ (9)

    and for prolate spheroidal coordinates defined by
    $\displaystyle x$ $\textstyle =$ $\displaystyle c \xi \sin \gamma \cos \phi$  
    $\displaystyle y$ $\textstyle =$ $\displaystyle c \xi \sin \gamma \sin \phi$  
    $\displaystyle z$ $\textstyle =$ $\displaystyle c \sqrt{1+\xi^2} \cos\gamma$ (10)

    The metric tensor $g_{ij}$ is defined to be
    \begin{displaymath}
g_{ij}= \sum_{k=1}^3
\frac{\partial x_k}{\partial u_i}
\frac{\partial x_k}{\partial u_j}
\end{displaymath} (11)

    where $x_1$, $x_2$, $x_3$ are $x$, $y$, and $z$, and $u_1$, $u_2$, and $u_3$ are $\xi$, $\eta$, and $\phi$. Explain why a diagonal metric tensor indicates an orthogonal coordinate system.

    b.
    The Laplacian can be derived by using the chain rule. For an orthogonal coordinate system, the Laplacian simplifies to
    \begin{displaymath}
\nabla^2 = \frac{1}{h_1 h_2 h_3} \sum_{\rm cyclic}
\frac{\pa...
... ( \frac{ h_j h_k}{h_i} \right )
\frac{\partial}{\partial u_i}
\end{displaymath} (12)

    where the cyclic sum means that $i,j,k$ take on the values $(1,2,3)$, $(2,3,1)$, and $(3,1,2)$, and $h_i \equiv \sqrt{g_{ii}}$. Similarly the volume element for volume integrals are $d^3r = h_1 h_2 h_3 du_1 du_2 du_3$ and the surface elements on surfaces of constant $u_i$ are $h_j h_k du_j du_k$. The various vector derivative operations are
    $\displaystyle \vec \nabla F$ $\textstyle =$ $\displaystyle \sum_i \frac{1}{h_i}
\frac{\partial}{\partial u_i} F
\hat u_i \,.$  
    $\displaystyle \vec \nabla \cdot \vec A$ $\textstyle =$ $\displaystyle \frac{1}{h_1 h_2 h_3} \sum_{\rm cyclic}
\frac{\partial}{\partial u_i} (h_j h_k A_i)$  
    $\displaystyle \vec \nabla \times \vec A$ $\textstyle =$ $\displaystyle \sum_{\rm cyclic}
\frac{1}{h_j h_k} \left [
\frac{\partial}{\partial u_j} (h_k A_k)
-\frac{\partial}{\partial u_k} (h_j A_j) \right ] \hat u_i$ (13)

    Show for oblate spheroidal coordinates we get
    \begin{displaymath}
\nabla^2 = \frac{1}{c^2 (\xi^2+\cos^2 \gamma)} \left \{
\fra...
...^2(\xi^2+1) \sin^2 \gamma}
\frac{\partial^2}{\partial \phi^2}
\end{displaymath} (14)

    and for prolate spheroidal coordinates
    \begin{displaymath}
\nabla^2 = \frac{\sqrt{1+\xi^2}}{c^2 (\xi^2+\sin^2 \gamma) \...
...1}{c^2 \xi^2 \sin^2 \gamma} \frac{\partial^2}{\partial \phi^2}
\end{displaymath} (15)

  5. Find values corresponding to constant $\xi$, $\gamma$, or $\phi$ along with the value of $c$ and either prolate or oblate spheroidal coordinates that correspond to
    a.
    A prolate spheroid, defined by the equation
    \begin{displaymath}
\frac{x^2+y^2}{b^2}+\frac{z^2}{a^2} = 1
\end{displaymath} (16)

    where $a > b$.
    b.
    A plane with a circular hole of radius $a$.
    c.
    A oblate spheroid, defined by the equation
    \begin{displaymath}
\frac{x^2+y^2}{a^2}+\frac{z^2}{b^2} = 1
\end{displaymath} (17)

    where $a > b$.
    d.
    A disk of radius $a$.

  6. If you know one solution to a second order linear ordinary differential equation, you can find the general solution by integration. To show this, assume that $y_1(x)$ is a nontrivial solution to
    \begin{displaymath}
\frac{d^2 y}{dx^2} + p(x) \frac{dy}{dx} + q(x) y = 0
\end{displaymath} (18)

    where $p(x)$ and $q(x)$ are arbitrary functions of $x$. Write the general solution as $y(x) = f(x)y_1(x)$, and by substituting into the the differential equation, show that it can be integrated to give the general form for $f(x)$ which gives the general solution. (Note, anyone who mentions or uses the Wronskian gets zero credit.)

    As a check, (but substituting this into the differential equation and showing it is the answer gets you zero credit), your result should be

    \begin{displaymath}
y(x) = A y_1(x)+B y_1(x) \int dx \frac{e^{-\int dx p(x)}}{y_1^2(x)}
\end{displaymath} (19)

    where $A$ and $B$ are arbitrary constants. Notice $q(x)$ does not appear.

    Verify that the method works by applying it to the ordinary Legendre equation

    \begin{displaymath}
\frac{dP}{dx} \left [ (1-x^2) \frac{dP}{dx} \right ] + \ell (\ell+1) P = 0
\end{displaymath} (20)

    for the case where $\ell = 1$, and the Legendre polynomial solution is $P_1(x) = x$, and compare your general solution to
    \begin{displaymath}
A P_1(x) + B Q_1(x)
\end{displaymath} (21)

    where $Q_1(x)$ is the Legendre function of the second kind (defined, for example, in Abramowtiz and Stegun 8.4.4).

  7. Far from a plane conducting surface at $z=0$ with a circular hole of radius $a$ centered at the origin, the electric field is $E_0 \hat z$ for large positive $z$ and zero for large negative $z$. Use oblate spheroidal coordinates and assuming, correctly, that the potential has the same angular, $\gamma, \phi$, dependence as a constant field in the $z$ direction, so that you can use the known solution for a constant field and the result of problem 6 to find the form of the general solution. Calculate the charge density on the top and bottom of the sheet as a function of $\rho$ the usual cylindrical radial distance. You can check your result by comparing with the result in Jackson's Problem 3.25 after converting it to Gaussian units.

    Note: Make sure your potential is continuous with continuous derivatives as you cross through the hole from positive to negative $z$.


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