406 Sir W. Thomson. On Electrostatic Screening [Apr. 



the eqnipotential surface c = ^a, exceeds in electrostatic capacity a 

 plane metal surface through the poles of the diagram (Plate XIII, 

 reproduced in 9 below), with the surroundings described in Art. 204, 

 and supplies the datum requisite for finding the exact amount of the 

 excess. The reason for the greatness of the excess clearly is that the 

 surface c = o, which just touches the plane through the poles of the 

 diagram midway between the poles, is everywhere nearer than this 

 plane to the other plate of the condenser. (See 7 below.) 



3. For c = a/6 we have, by (11) of Art. 205, a = o, and the corre- 

 sponding equi potential, partially shown in Maxwell's diagram, is a 

 set of curves concave towards z = oo , and asymptotic to the lines 

 a- = (t^)a, t denoting any integer, (See 10 to 13 below.) For 

 every value of c less than a/6, the equipotential is a row of ovals ; and 

 the grating formed by constructing these ovals in metal has less 

 electrostatic capacity in the circumstances described in Art. 205 than 

 a plane through the poles or the ovals (this being no doubt what is 

 meant by " a plane ... in the same position " as the grating). 



4. For every value of c exceeding a/6 the equipotential, instead of 

 being the boundary of a grating, is a continuous corrugated surface, 

 and its electrostatic capacity exceeds that of the plane through the 

 poles. 



5. Begin now afresh, and let it be required to find the electric force 

 in the air on either side of an infinite row of parallel bars at equal 

 consecutive distances, a, each uniformly charged with electricity. 

 Let pa be the quantity per unit length on each bar, so that p would 

 be the surface density, if the same quantity were uniformly dis- 

 tributed over the plane of the bars. Taking in one of the bars, 

 OX perpendicular to the bars, and OZ perpendicular to their plane, 

 we find (by Fourier's method) for the z-component of force at any 

 point (x, z) for which z is positive, 



Z = 



a 



where m 2w/a 



Summing this we find 



Z = 



_ 2irp 



a ** 2 cos mx + e 



This has equal positive and negative values for equal positive and 

 negative values of z, and it therefore gives the value of the s-force, 

 not only for positive, but also for negative, values of z. Taking now 

 f Zdz, with constant assigned to make the integral zero for z = + D, 

 we tirid 



V 



= pa (log - - ---- r mD ) (4) 



V fi ~ 2COS TH3- + 6-~ / ........ V/ 



