sphere. it a charge Q, which we will consider to be divided into a series of small charges 91, 92, 93, &c., distributed over the surface of the What is the potential of the sphere? Or, what will be the work required to bring our unit from infinity. up to the sphere? R Now let us consider what we mean by an isolated sphere.' There is, on the walls, &c., at a distance (let us suppose) of r centimètres from the centre O, a charge equal to - Q. Hence the true potential of O and of the sphere will be But if is so large as compared with R that the latter term. may be neglected, then we have V = Q This shows clearly the meaning of isolated; and also shows how, for the same charge, the potential will be smaller as becomes smaller, or as the walls close in (see Chapter V. § 8). § 20. Capacity of an Isolated Sphere.—By the definition of capacity given in Chapter V. § 4, we have for an isolated sphere Q Q = Hence it follows that R But we have just found that V KR. Or we arrive at the result that The capacity of an isolated sphere is measured numerically by its radius expressed in centimetres. Note. The reasoning just employed is of very general use. Suppose that we find by calculation that where / and / involve only one or more of the following quantities, viz. dimensions of the plates, distance between them, and nature of the dielectric. We may then argue that, by the definition of capacity given in Chapter V. § 4, and Chapter VI. § 4, it follows that K = /, or K=h, in the two cases respectively. $ 21. Distribution; from the Potential Point of View.-In hydrostatics it is frequently very useful to consider the distribution of a liquid to be such as will cause every part of the free surface to be at the same level. The more immediate cause of the distribution of a liquid is the action of gravitation-forces; and a level free surface rather follows from this. But still the former view is often convenient. So, in electrostatics, the cause of distribution is the direct force acting on each elementary portion, of the charge. But here also it is often convenient to regard the distribution on a conductor as such as will cause each part of it to be at the same electrostatic potential. This last must be the resulting condition, since otherwise we could not have equilibrium. We will take one case as an example. 'What must be the general distribution on an elongated cylinder that it may be all at one V ;' or, we might say, 'that it may take the same work to bring up our + unit to any part of it?' If the distribution were uniform, it would certainly require more work to bring up our unit to the central portion, into the presence of the whole charge, than to bring it up to the end, out of the way of most of the charge. But, if the charge be more crowded at the ends, we see how it may now require the same work in the latter case as in the former. Thus, the well-known distribution on an elongated cylinder is such as to cause the whole to be at one potential, though it more directly results from the equilibrium of the tangential components of the electrostatic forces. § 22. Two Spheres of Different Radii.-Let us consider two spheres of different radii, R, and R. (1) If they have the same charge Q, then = or = These results should be translated into words, and noted carefully by the student. § 23. Potential, and Density, distinguished.-By density of charge, we mean quantity of charge per unit area. We use the symbol p to signify the numerical value of density of charge. We have, by definition, where Q is the quantity on the area S; all of course in C.G.S. units. Now we have seen that, on any conductor but a sphere, the density is not constant all over, while the potential is. At a point the density may be very great, though the potential be small. We will illustrate the difference and the connection between density and potential by the case of spheres. (i.) Consider two spheres of equal radius R. Then, for each, are equal, and both V and p are proportional to Q. (ii.) But now consider two spheres of different radii R, and R2, and let them have the same potential V. Then for the two respectively we have Q1 = VR1; Q2 2 = = V R2; are not the same, and $ 24. Force on a + Unit, acting Perpendicularly to a Conducting Surface. It is for many purposes very important to know how great is the force on unit at a surface, urging it normally to the surface. The resultant force must evidently be normal, since any tangential force would imply that the conducting surface was not at one potential. We will take the case of a sphere. Let the diagram represent a sphere of radius R, and total charge Q; and let P be a point on the sphere. It is required to find the force at P acting on + unit. Now by the result of calculation, stated in § 17 (4), we learned that the action of the charged sphere on a point external to it is the same as if the whole charge Q were collected at the centre E. This is true quite up to the surface; or is true when we take our point P to be just on the external surface. Hence on a + unit at P there is acting a normal force equal to R E be the uniform density of charge over the sphere, then or Q = 4π R2 p. Hence, for the force F acting on + 4πp in dynes. Now if the radius of the sphere increase without limit, so that the surface at P becomes ultimately plane, no change will occur in the expression just found, provided that remains constant. Ρ Hence the formula is true for a plane surface on which the density is p. It is indeed true for any surface, p being the density at the point considered. This general result is, in more advanced Courses, proved independently, without reference to a sphere. $ 25. Important Case of a Spherical Condenser. We will now show how to find the capacity of a spherical condenser. As in the case of an isolated sphere we found two expressions for the potential of the sphere, and by equating them found the capacity of the sphere; so here, in the case of two spherical surfaces forming a condenser, we shall find two expressions for the difference of potentials of the two surfaces, and then by equating them shall find an expression for the capacity of the condenser. The figure represents the ideal spherical condenser, in which there are two concentric spherical conductors, the inner being totally enclosed by the outer. In the practical condenser we must have an opening through the outer conductor, and an insulated wire and knob connected with the inner; as in Faraday's condenser, Chapter IX. § 4. L Let C be the centre. Let the radius of the inner conductor be RA cms., and the charge on it be + Q; then there will be on the inside surface of the outer conductor a charge of - Q, by Chapter IV. § 16. Let the inner radius of this outer conductor be RB cms., and its outer radius be r cms.; and let there be on the outer surface of B a charge Q'. Α As we wish to find an expression for the difference of potential (VA - VB) between the two surfaces, we will find VA and Vв separately first. The reader must note that Q is the bound charge of Chapter VI. Now VA is the potential of any point on, or in, A, since A is a conductor containing no insulated charged bodies; and, therefore, VA is the potential at C. B And V1 will be the same as the potential of a point just on the outer surface of B. But, by § 17 (4), this will be the same as that which would be due to all the charges collected at the common centre C. as one expression for the difference of potential between A and B. But, by the definition of capacity, we have also that The quantity Q', and the outer radius r, have both disappeared from the expression. And this is what we should expect, since Q' has opposite to it on the walls, &c., a charge equal to Q', and these two charges with the intervening air as a dielectric form another |