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CYK/2007/PH 410/Quiz 1 Electrodynamics I

  1. A sphere of radius $ R$ carries a charge density $ \rho(r)=kr$ (where $ k$ is a constant). Find the energy of the configuration.[4]
  2. A point charge $ q$ of mass $ m$ is released from rest at a distance $ d$ from an infinite grounded conducting plane. How long will it take for the charge to hit the plane?[3]
  3. Consider a charge free volume $ V$ bounded by a closed surface $ S$ that consists of several separate surfaces (Conductors) $ S_i$ each held at potential $ V_i$ . Let $ \Psi(\v{x})$ be a well behaved function in $ V$ and on S, with a value equal to $ V_i$ on each surface $ S_i$ , but otherwise arbitraty for the present. Define the quantity
    $\displaystyle W[\Psi]=\frac{1}{8\pi}\int_V\vert\nabla \Psi\vert^2 d^3x$      

    Prove using variational calculus:
    $ W[\Psi]$ , which is nonnegative by definition, is stationary and an absolute minimum if and only if $ \Psi$ satisfies the Laplace equation in $ V$ and takes on the specified values $ V_i$ on the surfaces $ S_i$ .[3]
Answers
Answer 1
The electric field for the given charge density can be found by the Gauss law

$\displaystyle \v{E} = \left\{ \begin{array}{ll} \dfrac{kr^2}{4\epsilon_0 }\v{\hat{r}} & r<R \ \dfrac{kR^4}{4\epsilon_0 r^2}\v{\hat{r}} & r>R \end{array} \right.$ (1)

Hence the energy is given by
$\displaystyle W$ $\displaystyle =$ $\displaystyle \frac{\epsilon_0 }{2}\int_{\textrm{all space}} \vert\v{E}\vert^2  dv$ (2)
  $\displaystyle =$ $\displaystyle \frac{\pi k^2}{8\epsilon_0 }\left[ \int_0^R r^6 dr + \int_R^\infty \frac{R^8 dr}{r^2} \right]$ (3)
  $\displaystyle =$ $\displaystyle \frac{\pi k^2 R^7}{7\epsilon_0 }$ (4)

Alternatively
The Potential at any $ r < R$ is given by

$\displaystyle \phi(r) = \frac{k}{3\epsilon_0 \left(R^3-\frac{r^3}{4}\right)}$ (5)

And then the energy is given by
$\displaystyle W$ $\displaystyle =$ $\displaystyle \frac{1}{2}\int_0^R \rho(r)\phi(r) 4\pi r^2 dr$ (6)
  $\displaystyle =$ $\displaystyle \frac{4\pi k^2}{6\epsilon_0 }\left(R^3\int_0^R r^3  dr -\frac{1}{4}\int_0^R r^6  dr\right)$ (7)
  $\displaystyle =$ $\displaystyle \frac{\pi k^2 R^7}{7\epsilon_0 }$ (8)

Answer 2
From the method of images, the force on charge $ q$ is

$\displaystyle \v{F} = \frac{q^2(-\v{\hat{i}})}{4\pi\epsilon_0 (2x)^2}$ (9)

By Newton's Law,
$\displaystyle m \ddot{x}$ $\displaystyle =$ $\displaystyle -\frac{q^2}{4\pi\epsilon_0 (2x)^2}$ (10)
$\displaystyle \dot{x}$ $\displaystyle =$ $\displaystyle \left( \frac{q^2}{8\pi\epsilon_0 m}\right)^{1/2}\left(\frac{1}{x}-\frac{1}{d} \right)^{1/2}$ (11)

Integrating further, we get,

$\displaystyle T = \frac{\pi d}{q}(2\pi\epsilon_0 m d)^{1/2}$ (12)

Answer 3
For a small change in function $ \psi$ , the change in the functional $ W$ is given by (to the first order)
$\displaystyle \delta W$ $\displaystyle =$ $\displaystyle W[\psi+\delta\psi]-W[\psi]$ (13)
  $\displaystyle =$ $\displaystyle \frac{1}{4\pi}\oint_S \delta\psi \nabla\psi\cdot ds - \frac{1}{8\pi}\int \nabla^2\psi \delta\psi dv$ (14)

The surface integral vanishes, since $ \delta\psi=0$ on surface. Hence $ \delta W$ will vanish for all $ \delta\psi$ if and only if $ \nabla^2\psi=0$ .




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Charudatt Kadolkar 2007-02-11