Memoirs presented to the Cambridge philosophical society on the occasion of the jubilee of Sir George Gabriel Stokes, bart., Hon. LL. D., Hon. SC. D., Lucasian professor.

WITH APPLICATIONS TO ELECTROSTATIC PROBLEMS. 283
which is the same thing as
=pQ + - sin-l cos (O - 8) secha -p + -sin- cos0 + o)sech a............ (2),
where P' is the optical image of P in the disc. On putting in this expression (2), for
U, the values == r, 0 =- r, and remembering that over the disc PQ = P'Q, we verify
at once that U vanishes on both surfaces of the disc. If Q coincides with the point
(po, -00, 00) the function U remains finite.
The Green's function GpQ which is a function that is finite and continuous throughout
the whole of ordinary (the first) space, everywhere satisfies Laplace's equation, and is
equal to 1/PQ over both surfaces of the disc, is given by GpQ= P-U, hence the
required value of GPQ is
GPQY = PQ - sin-l cos (8 - 0) sech 2 a + P +- sin- {-cos ( + o) sech } a
=Q* -cos1 cos- (O(9-0) sech a- +/. - c oscos (O + 0o) sech.........(3),
PQ 7 r i 2 7r 2 2
the numerically smallest values, as before, of the inverse circular functions being taken.
It will be observed that in interpreting these formulae (2) and (3), the second copy of
space, having served its purpose, may be supposed to be removed.
THE DISTRIBUTION OF ELECTRICITY ON A CONDUCTING DISC UNDER THE INFLUENCE
OF A CHARGED POINT.
6. If we suppose a thin conducting disc to be placed in the position of the fundamental circle of the coordinate system, to be connected to earth, and influenced by a
charge q at the point P(p,, 0 b, ) on the positive side, the potential of the system at
any point Q is q U where U is given by (2), and the potential of the charge on the
disc is - q. GpQ. We shall now throw these potentials into a more geometrical form.
We have
sin-l {cos 2 (O - 0o) sech a 4} = tan-l'" ( -,)
1Cosh2 -a - cos2 ( - 00)
\/2 cos - (0 -,o)
=tan-' 2 }
=/cosh a - cos (O - 0o)
now take an auxiliary point L, of which the coordinates are po, 0 I-, o0, the upper
or lower sign being taken according as 0 is positive or negative (-rr < 0 < r). Thus L
and Q are always on opposite sides of the dise; using the formulae of Art. 1, we find
L- 2 a2 cos0 -2a'cos- 0
CL2 - au = a - _CQ -
cosh p, + cos 0 ' cosh p - cos 0 '
PL - f 1 + cos (0-0o) )t Çcosh p - cos 0\'
PQL cosh a - cos (0 - 0,) cosh po + cos Qj '
36-2