Page:Elementary Text-book of Physics (Anthony, 1897).djvu/321

§ 267] of force, we notice that if $$N$$ represent the number of tubes of force which pass through unit area drawn at a certain point in the field normal to the lines of force, and if $$s$$ represent a small area normal to the lines of force, then $$Ns$$ represents the number of tubes of force which pass through that area, and the tension on that area is $$\tfrac{1}{2}FNs.$$ Now we have seen that in any field in which the dielectric constant is $$K, F = \frac{4\pi N}{K}\cdot$$ Hence, substituting for $$N$$ and dividing by $$s,$$ we have the tension on unit area given by $$\frac{2\pi N^2}{K} = \frac{KF^2}{8\pi}\cdot$$ On the view which we are now taking it is natural to consider the work done in charging bodies in a field as expended in modifying the dielectric or in setting up unit tubes in it. We will examine on this supposition the distribution of energy in the electrical field. It has already been proved (§ 263) that the energy of a system of charged bodies is equal to $$\tfrac{1}{2} \textstyle \sum \displaystyle QV,$$ where $$Q$$ is the charge and $$V$$ the potential of each body. Let us consider a tube of force starting from a body at potential $$V_{1}$$ and proceeding to another body at potential $$V_{2};$$ the charges at the ends of these tubes of force are equal and of opposite sign. The energy of the first conductor due to the portion of its charge we are now considering, which may be called $$q,$$ is $$\tfrac{1}{2}qV_{1};$$ the energy of the second conductor due to the corresponding equal charge is $$\tfrac{1}{2} qV_{2}.$$ The energy, therefore, due to the charges associated with a tube of force is $$\tfrac{1}{2} q(V_{1} - V_{2}).$$ All charges in the field may be associated in this way in pairs, and the total energy of the field expressed by $$\tfrac{1}{2} \textstyle \sum \displaystyle q (V_{1} - V_{2}).$$ Now $$V_{1} - V_{2}$$ measures the work done by the electrical forces in moving a unit charge from the first conductor to the second. If $$V$$ represent the force at any point in a tube of force and $$\Delta l$$ an element of length of the tube, the product $$F \Delta l$$ represents the work done in moving the unit charge along that element, and the total work done in moving over the length of the whole tube is $$\textstyle \sum \displaystyle F \Delta l = V_{1} - V_{2}.$$ The energy associated with the whole tube is therefore $$\tfrac{1}{2}q \textstyle \sum \displaystyle F\Delta l,$$ and if we assume the unit length so small that the force does not appreciably vary within it, this energy may be considered as distributed