1913] TERMINATED RODS IN ELECTRICAL PROBLEMS 147
Thus altogether
(6)
For i/,,'2 we may write J^ ; so that if in (2) we take
_
we shall have a function which satisfies the necessary conditions, and at z = 0 assumes the value 1 from 0 to a and that expressed in (1) from a to 6. But the values of dVjdz are not the same on the two sides.
If we call the value, so determined on the positive as well as upon the negative side, V0, we may denote the true value of Vby V0+ V. The conditions for V will then be the satisfaction of Laplace's equation throughout the dielectric (except at z = 0), that on the negative side it make V' = Q both when r = a and when r = b, and vanish at z — — oo , and on the positive side V = 0 when r = b and when z = + oo , and that when z = 0 V assume the same value on the two sides between a and b and on the positive side the value zero from 0 to a. A further condition for the exact solution is that dV/dz, or dV0/dz + dV'/dz, shall be the same on the two sides from r = a to r = b when z = 0.
Now whatever may be in other respects the character of V on the negative side, it can be expressed by the series
V/ = Hl^(h1r)e^z + H^(h;,r)e^z+...} ............... (8)
where (f> (J^r), &c. are the normal functions appropriate to the symmetrical vibrations of an annular membrane of radii a and b, so that </> (hr) vanishes for r = a, r — b. In the usual notation we may write

~ J9(ka)
with the further condition
0, .................. (10)
determining the values of h. The function <£ satisfies the same differential equation as do J0 and YQ.
Considering for the present only one term of the series (8), we have to find for the positive side a function which shall satisfy the other necessary conditions and when z = 0 make V — 0 from 0 to a, and V = H<f> (hr) from a to b. As before, such a function may be expressed by
V' = BlJ0(klr)e^ + S2J0(k2r)e^ + ...) .............. (11)
and the only remaining question is to find the coefficients B. For this purpose we require to evaluate
<j> (hr) J"0 (kr) r dr.
b
10—2 K