Page:Elementary Text-book of Physics (Anthony, 1897).djvu/318

304 that these other conductors are at any distance from $$A,$$ and that they are at the common potential zero. They are then equivalent to a single conductor $$B$$ surrounding the conductor $$A.$$ Lines of force start from every point of $$A$$ and pass to corresponding points of $$B.$$ Mark out a small area on the surface $$A,$$ and consider the closed surface formed by the lines of force passing through the contour of that area and surfaces dravrn in the dielectric just outside the conductor $$A$$ and just outside the conductor $$B.$$ This closed surface is a tube of force, and if $$F_{A}$$ and $$F_{B}$$ represent the forces acting at the two cross-sections of the tube at $$A$$ and $$B$$ respectively, and $$s_{A}$$ and $$s_{B}$$ represent the areas of those cross-sections, we have (§ 56), $$F_{A}s_{A} = -F_{B}s_{B},$$ the forces being considered as directed along the normals drawn outward from the conductors. Since the force within the conductors vanishes, the force just outside the surface of $$A$$ is $$F_{A} = 4\pi \sigma_{A},$$ and that just outside the surface of $$B$$ is $$F_{B} = 4\pi \sigma_{B}.$$ Using these values for $$F_{A}$$ and $$F_{B},$$ we have $$\sigma_{A}s_{A} = -\sigma_{B}s_{B}.$$ Now these products are equal to the quantities of electricity present on the areas $$s_{A}$$ and $$s_{B},$$ so that we have $$q_{A} = -q_{B}.$$ The charges at the two ends of the tube of force are therefore equal and of opposite sign. Since the tubes of force which proceed from $$A$$ either extend to infinity or end on conductors, the charges on those conductors are never greater than the total charge on $$A.$$ If, as we have assumed, the conductors $$B$$ completely surround $$A,$$ the charges on $$B$$ are equal to the charge on $$A.$$ If we divide the surface of $$A$$ into areas upon each of which a unit charge of electricity is present, and erect tubes of force upon those areas, the dielectric will be mapped out by those tubes. Such a tube may be called a unit tube or a Faraday tube, in accordance with the proposition of J. J. Thomson.

266. Electrical Forces explained by Tubes of Force.—The strength of the field at any point in the dielectric is inversely as the area of the normal cross-section of the unit tube of force at that point. For, by § 56, the product Fs is constant throughout the tube. At the surface of the conductor from which the tube starts, $$F$$ is equal to $$4\pi \sigma$$ and $$Fs = 4\pi \sigma s = 4\pi,$$ since $$s$$ is the