Page:Elementary Text-book of Physics (Anthony, 1897).djvu/319

§ 366] of the unit tube at the charged surface, and $$\sigma s$$ is therefore equal to unity. The force $$F$$ at any point in the dielectric is therefore equal to $$\frac{4\pi}{s},$$ or is inversely as the cross-section of the tube. If we represent by $$N$$ the number of unit tubes which pass through a unit area of an equipotential surface, and if we assume that the force is appreciably constant over this area, we have $$N = \frac{1}{s}$$ and $$F = 4\pi N,$$ that is, the force at any point in the field is proportional to the number of unit tubes which pass perpendicularly through a unit area at that point.

In the discussion up to this point we have assumed that the medium between the two conductors has the specific inductive capacity or dielectric constant unity. If the dielectric constant be not unity, but some other number, say $$K,$$ the difference of potential between $$A$$ and $$B$$ that will be produced by a given charge on $$A$$ is less than that which will be produced when the dielectric constant is unity, in the ratio of 1 to $$K.$$ The general expression for the force in the field is therefore $$F = \frac{4\pi N}{K}\cdot$$

The electric pressure or force on unit area of the surface of the conductor, when the conductor is surrounded by a medium of which the dielectric constant is $$K,$$ is given by $$\frac{2\pi \sigma^2}{K}\cdot$$ This may be seen at once by applying the proof of § 256 to this case, remembering that, as has just been shown, the force outside the conductor is given by $$\frac{4\pi \sigma}{K}\cdot$$ Now on the view here taken, that the electrical forces are due to actions in the dielectric, this pressure should not be looked at as the result of the repulsions of the various elements of charge on the conductor, but rather as the result of some action in the dielectric. This action must be a pull or tension on the surface of the conductor. Since $$\sigma$$ represents the number of unit tubes which proceed from unit area of the conductor, this pull is equal to $$\frac{2\pi \sigma}{K}$$ applied to the end of each unit tube, or since the force