Hence the potential at the surface of the sphere, and therefore the potential of the sphere, is Q/R, where R is the radius of the sphere in centimetres. The quantity of electricity which must be given to the sphere to raise it to unit potential is therefore R electrostatic units. The capacity of a conductor is defined to be the charge required to raise its potential to unity, all other charged conductors being at an infinite distance. This capacity is then a function of the geometrical dimensions of the conductor, and can be mathematically determined in certain cases. Since the potential of a small charge of electricity dQ at a distance r is equal to dQ/r, and since the potential of all parts of a conductor is the same in those cases in which the distribution of surface density of electrification is uniform or symmetrical with respect to some point or axis in the conductor, we can calculate the potential by simply summing up terms like σdS/r, where dS is an element of surface, σ the surface density of electricity on it, and r the distance from the symmetrical centre. The capacity is then obtained as the quotient of the whole charge by this potential. Thus the distribution of electricity on a sphere in free space must be uniform, and all parts of the charge are at an Capacity of a sphere. equal distance R from the centre. Accordingly the potential at the centre is Q/R. But this must be the potential of the sphere, since all parts are at the same potential V. Since the capacity C is the ratio of charge to potential, the capacity of the sphere in free space is Q/V = R, or is numerically the same as its radius reckoned in centimetres.
We can thus easily calculate the capacity of a long thin wire like a telegraph wire far removed from the earth, as follows: Let 2r be the diameter of the wire, l its length, and σ the uniform Capacity of a thin rod. surface electric density. Then consider a thin annulus of the wire of width dx; the charge on it is equal to 2πrσ/dx units, and the potential V at a point on the axis at a distance x from the annulus due to this elementary charge is
| V = 2 ∫ l/2 0 | 2πrσ | dx = 4πrσ { loge(½l + √r² + ¼l²) − loger}. |
| √(r² + x²) |
If, then, r is small compared with l, we have V = 4πrσloge l/r. But the charge is Q = 2πrσ, and therefore the capacity of the thin wire is given by
C = ½ loge l/r
(2).
A more difficult case is presented by the ellipsoid[5]. We have first to determine the mode in which electricity distributes itself on a conducting ellipsoid in free space. It must be such a distribution that the potential in the interior will be Potential of an ellipsoid. constant, since the electric force must be zero. It is a well-known theorem in attractions that if a shell is made of gravitative matter whose inner and outer surfaces are similar ellipsoids, it exercises no attraction on a particle of matter in its interior[6]. Consider then an ellipsoidal shell the axes of whose bounding surfaces are (a, b, c) and (a + da), (b + db), (c + dc), where da/a = db/b = dc/c = μ. The potential of such a shell at any internal point is constant, and the equipotential surfaces for external space are ellipsoids confocal with the ellipsoidal shell. Hence if we distribute electricity over an ellipsoid, so that its density is everywhere proportional to the thickness of a shell formed by describing round the ellipsoid a similar and slightly larger one, that distribution will be in equilibrium and will produce a constant potential throughout the interior. Thus if σ is the surface density, δ the thickness of the shell at any point, and ρ the assumed volume density of the matter of the shell, we have σ = Aδρ. Then the quantity of electricity on any element of surface dS is A times the mass of the corresponding element of the shell; and if Q is the whole quantity of electricity on the ellipsoid, Q = A times the whole mass of the shell. This mass is equal to 4πabcρμ; therefore Q = A4πabcρμ and δ = μp, where p is the length of the perpendicular let fall from the centre of the ellipsoid on the tangent plane. Hence
σ = Qp / 4πabc
(3).
Accordingly for a given ellipsoid the surface density of free distribution of electricity on it is everywhere proportional to the length of the perpendicular let fall from the centre on Capacity of an ellipsoid. the tangent plane at that point. From this we can determine the capacity of the ellipsoid as follows: Let p be the length of the perpendicular from the centre of the ellipsoid, whose equation is x²/a² + y²/b² + z²/c² = 1 to the tangent plane at x, y, z. Then it can be shown that 1/p² = x²/a4 + y²/b4 + z²/c4 (see Frost’s Solid Geometry, p. 172). Hence the density σ is given by