Page:Elementary Text-book of Physics (Anthony, 1897).djvu/320

306 just outside the surface is equal to $$\frac{4\pi \sigma}{K},$$ the pull on the end of each unit tube is also given by $$\frac{F}{2}\cdot$$

The forces which act upon electrified bodies may therefore be considered as arising from tensions in the unit tubes, provided these tensions are not mechanically impossible. It may be shown that a medium in which such tensions exist is not in equilibrium unless pressures nuaaerically equal to the tensions, and at right angles to them, act throughout the medium. In order, therefore, that we may adequately represent the electrical field by the aid of unit tubes, we must assume a tension in each of these tubes along the lines of force and a pressure in every direction at right angles to it of the same numerical value. The tensions tend to shorten the tubes, the pressures to repel them from one another. All the forces which act between electrified bodies may be explained in terms of these actions between the tubes of force.

267. Energy in the Dielectric.—The tension on the cross-section of the unit tube at any point in the field is also $$\frac{F}{2}$$ where $$F$$ represents the force at that point. To show this, it is sufficient to suppose one of the equipotential surfaces around the charged body replaced by a conductor maintained at the potential of that surface. The distribution in the field between the two conductors will then be the same as before. By reasoning similar to that already employed, it is seen at once that the force on the surface of the new conductor which carries unit charge, or the pull on the end of a unit tube at that surface, is given by $$\frac{F}{2},$$ where $$F$$ is the force at a point in the end of the unit tube. No restriction has been made as to the particular equipotential surface chosen to be replaced by a conductor, and thus it appears that the tension or pull on the cross-section of the tube of force is everywhere equal to $$\frac{F}{2},$$ where $$F$$ is the force at a point in that cross-section.

To find the tension or pull across unit area normal to the lines