44 Electrostatics Field of Force [OH. n 



Lines of force can begin only on positive charges, and can end only on 

 negative charges. 



It is of course possible for a line of force to begin on a positive charge, 

 and go to infinity, the potential decreasing all the way, in which case the 

 line of force has, strictly speaking, no end at all. So also, a line of force may 

 come from infinity, and end on a negative charge. 



Obviously a line of force cannot begin and end on the same conductor, 

 for if it did so, the potential at its two ends would be the same. Hence there 

 can be no lines of force in the interior of a hollow conductor which contains 

 no charges ; consequently there can be no charges on its inner surface. 



Tubes of Force. 



55. Let us select any small area dS in the field, and let us draw the 

 lines of force through every point of the boundary of this small area. If 

 dS is taken sufficiently small, we can suppose the electric intensity to be the 

 same in magnitude and direction at every point of dS, so that the directions 

 of the lines of force at all the points on the boundary will be approximately 

 all parallel. By drawing the lines of force, then, we shall obtain a " tubular " 

 surface i.e., a surface such that in the neighbourhood of any point the 

 surface may be regarded as cylindrical. The surface obtained in this way 

 is called a " tube of force." A normal cross-section of a " tube of force " is a 

 section which cuts all the lines of force through its boundary at right angles. 

 It therefore forms part of an equipotential surface. 



56. THEOREM. If a>i, co 2 be the areas of two normal cross-sections of the 

 same tube of force, and R ly R 2 the intensities at these sections, then 



Consider the closed surface formed by the two cross-sections of areas 



&>!, ft> 2 , and of the part of the tube of force join- 

 ing them. There is no charge inside this surface, 



so that by Gauss' theorem, 1/^^ = 0. 



If the direction of the lines of force is from 

 G>! to &> 2 , then the outward normal intensity 

 FIG. 12. over W 2 is R z , so that the contribution from this 



area to the surface integral is R. 2 a) 2 . So also 



over (Wi the outward normal intensity is R 1} so that G^ gives a contribution 

 R^. Over the rest of the surface, the outward normal is perpendicular 

 to the electric intensity, so that N=Q, and this part of the surface con- 



tributes nothing to uNd8. The whole value of this integral, then, is 



R z a) 2 R l w l , 

 and since this, as we have seen, must vanish, the theorem is proved. 



