Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/404

simplified, for if $k$ is the specific resistance per unit of volume, then

$u=-{\frac {1}{k}}{\frac {dV}{dx}},\quad \quad v=-{\frac {1}{k}},\quad \quad w=-{\frac {1}{k}}{\frac {dV}{dz}},$

(13)

and if $\nu$ is the normal drawn at any point of the surface of separation from the first medium towards the second, the conduction of continuity is

${\frac {1}{k_{1}}}{\frac {dV_{1}}{d\nu }}={\frac {1}{k_{2}}}{\frac {dV_{2}}{d\nu }}.$

(14)

If $\theta _{1}$ and $\theta _{2}$ are the angles which the lines of flow in the first and second media respectively make with the normal to the surface of separation, then the tangents to these lines of flow are in the same plane with the normal and on opposite sides of it, and

$k_{1}\tan \theta _{1}=k_{2}\tan \theta _{2}.$

(15)

This may be called the law of refraction of lines of flow.

311.] As an example of the conditions which must be fulfilled when electricity crosses the surface of separation of two media, let us suppose the surface spherical and of radius $a$, the specific resistance being $k_{1}$ within and $k_{2}$ without the surface.

Let the potential, both within and without the surface, be expanded in solid harmonics, and let the part which depends on the surface harmonic $S_{i}$ be

$V_{1}=\left(A_{1}r^{i}+B_{1}r^{-(i+1)}\right),$

(1)

$V_{2}=\left(A_{2}r^{i}+B_{2}r^{-(i+1)}\right)S_{i}$

(2)

within and without the sphere respectively.

At the surface of separation where $r=a$ we must have

$V_{1}=V_{2}\quad \quad {\mbox{and}}\quad \quad {\frac {1}{k_{1}}}{\frac {dV_{1}}{dr}}={\frac {1}{k_{2}}}{\frac {dV_{2}}{dr}}.$

(3)

From these conditions we get the equations

$\left.{\begin{array}{l}(A_{1}-A_{2})a^{2i+1}+B_{1}-B_{2}=0,\\\left({\frac {1}{k_{1}}}A_{1}-{\frac {1}{k_{2}}}A_{2}\right)ia^{2i+1}-\left({\frac {1}{k_{1}}}B_{1}-{\frac {1}{k_{2}}}B_{2}\right)(i+1)=0.\end{array}}\right\}$

(4)

These equations are sufficient, when we know two of the four quantities $A_{1},\,A_{2},\,B_{1},\,B_{2},$ to deduce the other two.

Let us suppose $A_{1}$ and $B_{1}$ known, then we find the following expressions for $A_{2}$ and $B_{2}$,

$\left.{\begin{array}{l}A_{2}={\frac {(k_{1}(i+1)+k_{2}i)A_{1}+(k_{1}-k_{2})(i+1)B_{1}a^{-(2i+1)}}{k_{1}(2i+1)}},\\\\B_{2}={\frac {(k_{1}-k_{2})iA_{1}a^{2i+1}+(k_{1}i+k_{2}(i+1))B_{1}}{k_{1}(2i+1)}}.\end{array}}\right\}$

(5)