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PHY531 Problem Set 4. Due October 16, 2000

  1. An incompressible conducting charged liquid drop is initially spherical with a radius $a_0$ and has a total charge $Q$. The surface is then slightly perturbed. The distance from the origin to the surface as a function of the usual polar angles is
R(\theta, \phi) = a [ 1 + {\sum_{\ell m}}' \alpha_{\ell m} Y_{\ell m}
(\theta, \phi) ]  .
\end{displaymath} (1)

    The prime on the sum indicates that the $\ell=m=0$ term is omitted. The $\alpha_{\ell m}$ have magnitudes much less than 1, $a$ is a constant that takes the necessary value so that the volume remains $4\pi a_0^3/3$, and $R(\theta, \phi)$ is, of course, real.

    Show that the electrostatic energy and the surface area of the drop, correct to second order in $\alpha_{\ell m}$ are
    $\displaystyle W$ $\textstyle =$ $\displaystyle \frac{Q^2}{2a_0} \left [ 1 - \frac{1}{4\pi} {\sum_{\ell m} }'
(\ell-1) \vert\alpha_{\ell m}\vert^2 \right ]$  
    $\displaystyle A$ $\textstyle =$ $\displaystyle 4 \pi a_0^2\left [ 1 + \frac{1}{8\pi} {\sum_{\ell m} }'
(\ell-1) (\ell+2) \vert\alpha_{\ell m}\vert^2 \right ]  .$ (2)

    The total potential energy from a small distortion is
E_{\rm Total} = W + \gamma A
\end{displaymath} (3)

    where $\gamma$ is the surface tension of the liquid. Find the range of charge values that can be found on a stable spherical drop of radius $a_0$.

    A raindrop can be considered to be a conductor for electrostatic purposes. What is the potential difference, in Gaussian units, between the drop surface and infinity for a spherical drop that is just stable if its diameter is 1 millimeter? Convert this result to Volts. The surface tension of water at 20$^\circ$C is 72.75 erg/cm$^2$ from the American Insitute of Physics Handbook, third edition, Ed. D.E. Gray, (McGraw Hill, New York, 1972).

    This problem was first solved by Rayleigh in 1882.

  2. Jackson Problem 3.23

  3. a.
    Show that if $\Phi_1(r,\theta,\phi)$ is a solution of Laplace's equation, that
\Phi_2(r,\theta,\phi) = \frac{a}{r} \Phi_1
\left ( \frac{a^2}{r},\theta,\phi \right )
\end{displaymath} (4)

    is also a solution of Laplace's equation. This is called inversion on a sphere of radius $a$.

    Apply the result of part a to the potential of a point charge on the axis of an infinitely long grounded conducting circular cylinder, to show that the self capacitance of a conducting surface generated by rotating a circle of diameter $a$ about one of its tangents is
C = 2a \sum_{k=1}^\infty \frac{1}{J_1(\mu_k)}
\int_0^\infty e^{-\mu_k \sinh \phi} d\phi
\end{displaymath} (5)

    where $J_0(\mu_k) = 0$.

    Use the result of Jackson Problem 1.20 to give crude upper and lower bounds to the coefficient of $a$ in the self capacitance result.

    Numerically evaluate the coefficient of $a$ in the self capacitance to at least 3 place accuracy. You may find Euler's transformation, which speeds up the convergence of well behaved alternating series,
\sum_{m=0}^\infty (-1)^m u_m =
\sum_{m=0}^{n-1} (-1)^m u_m +\sum_{p=0}^\infty \frac{(-1)^p}{2^{p+1}}
\Delta^p u_n
\end{displaymath} (6)

    useful. Here $\Delta$ is the forward difference operator (i.e. $\Delta u_n = u_{n+1}-u_n$).

  4. Jackson Problem 3.3

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