Page:Motion of Electrification through a Dielectric.djvu/14

But if u=v in (40), the result is zero, unless ν₁=1, when we have the result (41). But if P be still further to the left, we shall have to add to (41) the solution due to the electrification which is ahead of P. So when the line is infinitely long both ways, we have double the result in (41), with independence of u/v again. But should q be a function of z, we do not have independence of u/v except in the already considered case of u=v, with plane waves, and no component of electric force parallel to the line of motion.

A Charged Straight Line moving Transversely.

17. Next, let the electrified line be in steady motion perpendicularly to its length. Let q be the linear density (constant), the z-axis that of the motion, the x-axis coincident with the electrified line and that of y upward on the paper. Then the A at P will be

$\scriptstyle {A={\frac {qu}{(1-u^{2}/v^{2})}}\log {\frac {x_{1}+\left\{x_{1}^{2}+y^{2}+z^{2}(1-u^{2}/v^{2})^{-1}\right\}^{\frac {1}{2}}}{x_{2}+\left\{x_{2}^{2}+y^{2}+z^{2}(1-u^{2}/v^{2})^{-1}\right\}^{\frac {1}{2}}}};\quad \cdots \cdots \cdots \cdots {\text{(42)}}}$

where y and z belong to P, and x₁, x₂ are the limiting values of x in the charged line. From this derive the solution in the case of an infinitely long line. It is

$\scriptstyle {cE={\frac {2q}{r}}\cdot {\frac {(1-u^{2}/v^{2})^{\frac {1}{2}}}{1-\nu ^{2}u^{2}/v^{2}}},\qquad H=cEuv,\quad \cdots \cdots \cdots \cdots \cdots \cdots {\text{(43)}}}$

where ν=sin θ understanding that E is radial, or along qP in the figure, and H rectilinear, parallel to the charged line. Terminating the fields internally at r=a, we have the case of a perfectly conducting cylinder of radius a, charged with q per unit of length, moving transversely. When u=v there is disappearance of E and H everywhere except in the plane $\scriptstyle {\theta ={\frac {1}{2}}\pi }$, as in the case of the sphere, and consequent infinite values. It is the curvature that permits this to occur, i.e. producing infinite values; of course it is the self-induction that is the cause of the conversion to a plane wave, here and in the other cases. There is some similarity between (43) and (29). In fact, (43) is the bidimensional equivalent of (29).

A Charged Plane moving Transversely.

18. Coming next to a plane distribution of electrification, let q be the surface-density, and the plane be moving perpendicularly to itself. Let it be of finite breadth and of infinite length, so that we may calculate H from (43). The result at P is

$\scriptstyle {H={\frac {qu}{(1-u^{2}/v^{2})^{\frac {1}{2}}}}\log {\frac {r_{1}^{2}-y_{1}^{2}u^{2}/v^{2}}{r_{2}^{2}-y_{2}^{2}u^{2}/v^{2}}}.\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(44)}}}$