Surface charge density of a conducting disk from that of a conducting ellipsoid

A basic result from applying Gauss’ Law of electromagnetism to a conductor is that any excess charge always ends up distributed entirely on the surface of the conductor. Furthermore, the excess charge will always distribute itself there so as to produce an equipotential surface. The resulting surface charge density distribution per unit area is usually difficult to obtain as a formula in closed form but it happens to be known for a conducting ellipsoid with general equation

$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \qquad \qquad \qquad \qquad \qquad (1)$

If the total excess charge on the surface of this conducting ellipsoid is $Q$ , the distribution of surface charge density per unit area is given by

$\sigma = \frac{Q}{4 \pi a b c } \big(\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}\big)^{-1/2} \qquad \qquad \qquad \qquad \qquad (2)$

(see, e.g., D. Griffiths, 1999, Introduction to Electrodynamics, Problem 2.52, page 109). Interestingly, it is possible to deduce from this the surface charge density distribution of other conducting solids that can be obtained as special cases of a conducting ellipsoid by suitable choices for the parameters $a$ , $b$ and $c$ , and I wanted to make a note of this here.

For example, we can immediately obtain the surface charge density distribution of a conducting sphere of radius $R$ by setting $a = b = c = R$ in (1) and (2) above to get the sphere

$x^2 + y^2 + z^2 = R^2 \qquad \qquad \qquad \qquad \qquad (3)$

and its corresponding surface charge density

$\sigma = \frac{Q}{4 \pi R^3 } \big(\frac{x^2}{R^4} + \frac{y^2}{R^4} + \frac{z^2}{R^4}\big)^{-1/2} = \frac{Q}{4 \pi} \bigg(R^6 \big(\frac{x^2}{R^4} + \frac{y^2}{R^4} + \frac{z^2}{R^4}\big)\bigg)^{-1/2}$

$= \frac{Q}{4 \pi} \big(R^2 (x^2 + y^2 + z^2)\big)^{-1/2} = \frac{Q}{4 \pi R^2} \qquad \qquad \qquad \qquad \qquad (4)$

where the last equality follows by (3).

Note that a unit sphere

$\tilde{x}^2 + \tilde{y}^2 + \tilde{z}^2 = 1 \qquad \qquad \qquad \qquad \qquad (5)$

can be transformed into the ellipsoid in (1) above via a scaling of each of the variables, i.e., the change of variables $x = a \tilde{x}$ , $y = b \tilde{y}$ , $z = c \tilde{z}$ . Thus, we can obtain the surface charge density of a circular conducting disk of radius $R$ coplanar with the $xy$ -plane and with centre at the origin by setting $c = 0$ and $a = b = R$ in (2), and using the fact that on the surface of the ellipsoid we have

$\frac{z^2}{c^2} = 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} \qquad \qquad \qquad \qquad \qquad (6)$

so that (2) can be rearranged to read

$\sigma = \frac{Q}{4 \pi ab} \bigg(c^2 \big(\frac{x^2}{a^2} + \frac{y^2}{b^2}\big) + 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}\bigg)^{-1/2} \qquad \qquad \qquad \qquad \qquad (7)$

Thus, setting $c = 0$ and $a = b = R$ in (7), we get the following formula for the surface charge density per unit area of a circular conducting disk of radius $R$ and total excess charge $Q$ (spread over both sides) at a point a radial distance $r = \sqrt{x^2 + y^2}$ from its centre:

$\sigma = \frac{Q}{4 \pi R^2} \big(1 - \frac{x^2}{R^2} - \frac{y^2}{R^2}\big)^{-1/2} = \frac{Q}{4 \pi R} \big(R^2 - r^2\big)^{-1/2} \qquad \qquad \qquad \qquad \qquad (8)$

Surface charge density of a conducting disk from that of a conducting ellipsoid

Published by Dr Christian P. H. Salas

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