If we define the positive direction along a tube of electric force as the direction in which a small body charged with positive electricity would tend to move, we can summarize the above facts in a simple form by saying that, _if we have any closed surface described in any manner in an electric field, the excess of the number of unit tubes which leave the surface over those which enter it is equal to 4[pi]-times the algebraic sum of all the electricity included within the surface_. Every tube of electric force must therefore begin and end on electrified surfaces of opposite sign, and the quantities of positive and negative electricity on its two ends are equal, since the force E just outside an electrified surface is normal to it and equal to [sigma]/4[pi], where [sigma] is the surface density; and since we have just proved that for the ends of a tube of force EdS = E¹dS', it follows that [sigma]dS = [sigma]'dS', or Q = Q', where Q and Q' are the quantities of electricity on the ends of the tube of force. Accordingly, since every tube sent out from a charged conductor must end somewhere on another charge of opposite sign, it follows that the two electricities always exist in equal quantity, and that it is impossible to create any quantity of one kind without creating an equal quantity of the opposite sign. [Illustration: FIG. 4.] We have next to consider the energy storage which takes place when electric charge is created, i.e. when the dielectric is strained or polarized. Since the potential of a conductor is defined to be the work required to move a unit of positive electricity from the surface of the earth or from an infinite distance from all electricity to the surface of the conductor, it follows that the work done in putting a small charge dq into a conductor at a potential v is v dq. Let us then suppose that a conductor originally at zero potential has its potential raised by administering to it small successive doses of electricity dq. The first raises its potential to v, the second to v' and so on, and the nth to V. Take any horizontal line and divide it into small elements of length each representing dq, and draw vertical lines representing the potentials v, v', &c., and after each dose. Since the potential rises proportionately to the quantity in the conductor, the ends of these ordinates will lie on a straight line and define a triangle whose base line is a length equal to the total quantity Q and height a length equal to the final potential V. The element of work done in introducing the quantity of electricity dq at a potential v is represented by the element of area of this triangle (see fig. 4), and hence the work done in charging the conductor with quantity Q to final potential V is ½QV, or since Q = CV, where C is its capacity, the work done is represented by ½CV² or by ½Q²/C.