An introduction to the mathematical theory of attraction ...

Examples. 235
potential in external space are found then as in Ex. 5, Art. 119. The totalmass
Ea on the spherical surface whose radius is a is given by the equation
Ea= - { OI - OJ1 + OI3 + OJ3 + &c. + GI1 + GJ1 - G13+ GJ3 + &c.
-OI2- J2 - J OI - OJ4 - &c. - GI2 - GJ2 - GI4- G - &c.},
where GJ\, GI2, &c. are to be taken as positive or negative according as the
points Ji, I2, &c. are on the same side of G as I1 or on the opposite side.
2. A spherical bowl is at potential zero under the influence of an external
electrified point O situated on the surface of the sphere (A) in space of which
geometrically the bowl forms a part; find the distribution of mass on the surface
of the bowl.
Invert from 0, the sphere (A) becomes a plane, and the plane base of the
bowl a sphere, so that the surface of the bowl is inverted into a plane circular
disk. If we suppose this disk at constant potential L', by Ex. 7, Art. 38, the
density o' of the distribution at any point P of the disk is given by the equation
V(a'2 - r'2)'
where k is a constant, a is the radius of the disk, and r' the distance of P' from
its centre; then, if S' be one surface of the disk,
Z^2f^ =4^r-',a' drP;
J J0 V('2 - r'2)
whence
' 1
27,r V/ t-'P'. P'K'
where H' and K' are the extremities of any chord of the disk passing through P'.
If H and K be the points on the edge of the bowl inverse to t' and K', we
have
HP OP OP. 0f PK OP OP. OK / R \3.H'P' 01 OH1 1-R2 'FK O'OK' - 2; a
StF:=:ml=::~~2_K 0 - - 'O = / --- —; also <-' [OP);
whence
L' xR /O. OK
2r O2P2JSH P
If e be the charge at 0, in order that the bowl should be at potential zero
by 2~, (b), Art 119, we have L'R = - e, and therefore,
-e 1 l OH.O OK
2=r2 OP2 sPH PK'