Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/406

364 then {{numb form| $$ \left. \begin{array}{l}A_1 = k_1k_2(2i+1)^2CA_3, \\ A_2=k_2(2i+1)(k_1(i+1)+k_2i)CA_3, \\ B_2=k_2i(2i+1)(k_1-k_2)a_1^{2i+1}CA_3, \\ B_3=i(k_2-k_1)(k_1(i+1)+k_2i)(a_2^{2i+1}-a_1^{2i+1})CA_3. \end{array} \right \} $$ | (7)}}

The difference between $$A_3$$ the undisturbed coefficient, and $$A_1$$ its value in the hollow within the spherical shell, is

Since this quantity is always positive whatever be the values of $$k_1$$ and $$k_2,$$ it follows that, whether the spherical shell conducts better or worse than the rest of the medium, the electrical action within the shell is less than it would otherwise be. If the shell is a better conductor than the rest of the medium it tends to equalize the potential all round the inner sphere. If it is a worse conductor, it tends to prevent the electrical currents from reaching the inner sphere at all.

The case of a solid sphere may be deduced from this by making $$a_1=0,$$ or it may be worked out independently.

313.] The most important term in the harmonic expansion is that in which $$i = 1,$$ for which {{numb form| $$ \left. \begin{array}{c} C= \frac{{1}}{{9k_1k_2+2(k_1-k_2)^2\left(1-(\frac{{a_1}}{{a_2}})^3\right)}}, \\ A_1=9k_1k_2CA_3, \quad \quad A_2=3k_2(2k_1+k_2)CA_3, \\ B_2=3k_2(k_1-k_2)a_1^3CA_3, \quad B_3=(k_2-k_1)(2k_1+k_2)(a_2^3-a_1^3)CA_3. \end{array}\right\} $$ |(9)}}

The case of a solid sphere of resistance $$k_2$$ may be deduced from this by making $$a_1 = 0.$$ We then have {{numb form| $$ \left .\begin{array}{l}A_2=\frac{{3k_2}}{{k_1+2k_2}}A_3, \quad \quad B_2=0, \\ B_3=\frac{{k_2-k_1}}{{k_1+2k_2}}a_2^3A_3. \end{array} \right \} $$ | (10)}}

It is easy to shew from the general expressions that the value of $$ B_3$$ in the case of a hollow sphere having a nucleus of resistance $$k_1$$ surrounded by a shell of resistance $$k_2,$$ is the same as that of a uniform solid sphere of the radius of the outer surface, and of resistance $$K$$, where