We have then a very important theorem as follows:—If any closed surface be described in an electric field which wholly encloses or wholly excludes electrified bodies, then the total flux through this surface is equal to 4π- times the total quantity of electricity within it.[16] This is commonly called Stokes’s theorem. The proof is as follows:—Consider any point-charge E of electricity included in any surface S, S, S (see fig. 3), and describe through it as centre a cone of small solid angle dω cutting out of the enclosing surface in two small areas dS and dS′ at distances x and x′. Then the electric force due to the point charge q at distance x is q/x, and the resolved part normal to the element of surface dS is q cosθ / x². The normal section of the cone at that point is equal to dS cosθ, and the solid angle dω is equal to dS cosθ / x². Hence the flux through dS is qdω. Accordingly, since the total solid angle round a point is 4π, it follows that the total flux through the closed surface due to the single point charge q is 4πq, and what is true for one point charge is true for any collection forming a total charge Q of any form. Hence the total electric flux due to a charge Q through an enclosing surface is 4πQ, and therefore is zero through one enclosing no electricity.
Stokes’s theorem becomes an obvious truism if applied to an incompressible fluid. Let a source of fluid be a point from which an incompressible fluid is emitted in all directions. Close to the source the stream lines will be radial lines. Let a very small sphere be described round the source, and let the strength of the source be defined as the total flow per second through the surface of this small sphere. Then if we have any number of sources enclosed by any surface, the total flow per second through this surface is equal to the total strengths of all the sources. If, however, we defined the strength of the source by the statement that the strength divided by the square of the distance gives the velocity of the liquid at that point, then the total flux through any enclosing surface would be 4π times the strengths of all the sources enclosed. To every proposition in electrostatics there is thus a corresponding one in the hydrokinetic theory of incompressible liquids.
Let us apply the above theorem to the case of a small parallel-epipedon or rectangular prism having sides dx, dy, dz respectively, its centre having co-ordinates (x, y, z). Its angular points have then co-ordinates (x ± ½dx, y ± ½dy, z ± ½dz). Let this rectangular prism be supposed to be wholly filled up with electricity of density ρ; then the total quantity in it is ρ dx dy dz. Consider the two faces perpendicular to the x-axis. Let V be the potential at the centre of the prism, then the normal forces on the two faces of area dy·dx are respectively
| − ( | dV | + ½ | d²V | dx) and ( | dV | − ½ | d²V | dx), |
| dx | dx² | dx | dx² |
and similar expressions for the normal forces to the other pairs of faces dx·dy, dz·dx. Hence, multiplying these normal forces by the areas of the corresponding faces, we have the total flux parallel to the x-axis given by −(d²V / dx²) dx dy dz, and similar expressions for the other sides. Hence the total flux is
| − ( | d²V | + | d²V | + | d²V | ) dx dy dz, |
| dx² | dy² | dz² |
and by the previous theorem this must be equal to 4πρdx dy dz.
Hence
| d²V | + | d²V | + | d²V | + 4πρ = 0 |
| dx² | dy² | dz² |
(18).