Page:MortonCharge.djvu/7

Showing that, as Mr. Heaviside pointed out, the field of a point charge is given when the conductor is an oblate spheroid whose axes have the ratio 1:k.

For a sphere the integral becomes

$\phi ={\frac {e}{8\pi k'a}}\log {\frac {\theta +k'a}{\theta -k'a}}$

where

$k'={\sqrt {1-k^{2}}}={\frac {u}{V}}$,

and θ is given by

${\frac {k^{2}(x^{2}+y^{2})}{\theta ^{2}-k'^{2}a^{2}}}+{\frac {z^{2}}{\theta ^{2}}}=1$.

To test the value of φ let us make k' approach zero, i. e. the motion becomes infinitely slow. θ is then =r.

Then

$\phi ={\frac {e}{8\pi a}}{\underset {k'=0}{Lt}}\log {\frac {(r+k'a)-\log(r-k'a)}{k'}}$ $={\frac {e}{8\pi a}}\cdot {\frac {2a}{r}}={\frac {e}{4\pi r}}$.

8. The mutual energy of a moving charge and external magnetic field has been given by Mr. Heaviside for the case of motion which is very slow compared with the velocity of radiation. It is eu A · cos (uA), where A is the circuital vector potential of the external field. Mr. Larmor, in the second part of his "Dynamical Theory" (Phil. Trans. 1895, p. 717), concludes that the same expression holds good for motion at any speed. He seems, however, to overlook the fact that in the general case the displacement-currents in the medium — being no longer derivable from a potential function — will make their appearance in the result as well as the convection-current eu.

If (F G H) is the vector potential, the part of the energy corresponding to the displacement currents will be

$\int (F{\dot {f}}+G{\dot {g}}+H{\dot {h}})d\tau$,

which in the case we have been considering becomes

$-u\int (F{\frac {df}{dz}}+G{\frac {dg}{dz}}+H{\frac {dh}{dz}})d\tau$

$=u\int (F{\frac {d^{2}\phi }{dx\ dz}}+G{\frac {d^{2}\phi }{dy\ dz}}+k^{2}H{\frac {d^{2}\phi }{dz^{2}}})d\tau$.