Then the potential at any point lying without the outer bounding surface will be For this expression, when substituted for V, satisfies the equation V2 V=0; it also agrees with the given value of the potential for every point of the axis, lying without the outer bounding surface, and it does not become infinite at any point within that surface. By the introduction of the expressions for zonal harmonics in the form of definite integrals, it will be found that if the value of either of these potentials for any point in the axis be denoted by (), the corresponding value for any other point, which can be reached without passing through any portion of the attracting mass, will be 3. We may next shew how to obtain, in terms of a series of zonal harmonics, an expression for the solid angle subtended by a circle at any point. We must first prove the following theorem. The solid angle, subtended by a closed plane curve at any point, is proportional to the component attraction perpendicular to the plane of the curve, exercised upon the point by a lamina, of uniform density and thickness, bounded by the closed plane curve. For, if ds be any element of such a lamina, r its distance from the attracted point, the inclination of r to the line perpendicular to the plane of the lamina, the elementary solid angle subtended by dS at the point will be dS cos 0 And the component attraction of the element of the lamina corresponding to dS in the direction perpendicular to its plane will be p being the density of the lamina, k its thickness. Hence, for this element, the component attraction is to the solid angle as pk to 1, and the same relation holding for every element of the lamina, we see that the component attraction of the whole lamina is to the solid angle subtended by the whole curve as pk to 1. Now, if the plane of xy be taken parallel to the plane of the lamina, and V be the potential of the lamina, its component attraction perpendicular to its plane will be dV Now since V is a potential we have ▼aV=0, whence dz d V*V=0, or V2 dz dv IV dz and satisfies all the analytical conditions to which a potential is subject. It follows that, if the solid angle subtended by a closed plane curve at any point (x, y, z) be denoted by w, w will be a function of x, y, z, satisfying the equation vw= 0. Hence, if the closed plane curve be a circle it follows that the magnitude of the solid angle which it subtends at any point may be obtained by first determining the solid angle which it subtends at any point of a line. drawn through its centre perpendicular to its plane, and then deducing the general expression by the employment of zonal harmonics. =0. Hence is itself a potential, Now let O be the centre of the circle, Q any point on the line drawn through O perpendicular to the plane of the circle, E any point in the circumference of the circle. With centre Q, and radius Q0, describe a circle, cutting QE in L. From L draw LN, perpendicular to QO. Let OE=c, 0Q=z. And the solid angle subtended by the circle at O To obtain the general expression for the solid angle subtended at any point, distant r from the centre, we first develope this expression in a converging series, proceeding by powers of z. This will be Hence, by similar reasoning to that already employed, we get, for the solid angle subtended at a point distant r from the centre, 4. We may deduce from this, expressions for the potential of a circular lamina, of uniform thickness and density, at an external point. For we see that, if V be the potential of such a lamina, k its thickness, and p its density, we have for a point on the axis, Hence we obtain the following expressions for the potential of an uniform circular lamina at a point distant r from the centre of the lamina : 1 P2 _1.1 P + ... 2 2.4 c3 It may be shewn that the solid angle may be expressed in the form 2π-2 • [c2 + {z + (22 — 2)3 cos 0}] and the potential of the lamina in the form 10. Mr c2 5. As another example, let it be required to determine the potential of a solid sphere, whose density varies inversely as the fifth power of the distance from a given external point O at any point of its mass. It is proved by the method of inversion (see Thomson and Tait's Natural Philosophy, Vol. 1, Art. 518) that the potential at any external point P' will be equal to M , O'P' Ο' being the image of O in the surface of the sphere, and M the mass of the sphere. We shall avail ourselves of this result to determine the potential at a given internal point. Let C be the centre of the sphere, O the given external point. Join CO, and let it cut the surface of the sphere in A, and in CA take a point O', such that CO. CƠ = ČA3. Then O' is the image of 0. Let P be any point in the body of the sphere, then we wish to find the potential of the sphere at P. Take O as pole, and OC as prime radius, let OP=r, POC 0. Also let CA= a, CO = c. = Let the density of the sphere at its centre be p, then its |