The Atom and the Charge of Electricity carried by it. 511 

 value of the integral in (4) or 



f 



2-40483 



xf{x) J (x) .^ = 0*6117, 



and hence by (4), as ^(2-40483) =0519, 



Al= (0-51^(2-405) 2 =0 ' 7852 - 



To find A 2 . The 25 equidistant ordinates of the original 

 curve ( straight line) are the ordinates of points in the curve 

 MBLBBP (fig. 2), the abscissae being taken from Table II. 

 We found it convenient to multiply the abscissae by 5 and 

 represent in inches. Our ordinates were multiplied by 10 and 

 represented in inches. The actual area was 2*36 square 

 inches and one-fiftieth of this, or O0472, is the value of the 

 integral in (4), 



Hence, as J 1 (5-5201) = -3403 5 



A 2x0-0472 _ ftWQ 



2 ~ (0-3403) 2 x (5-5201) 2 



We need not show the curves used in finding A 3 and A 4 . 

 The area of fig. 2 is the positive area MLNOM minus the 

 area LQPNL, but one need not think about whether an area 

 is positive or negative. It is only necessary to start the 

 planimeter-tracer from the point numbered in every case, 

 and go from to M and along the curve in the direction of 

 the increasing numbers to 24, then along the axes of abscissae, 

 ending at the point from which we started. 



XLVIII. The Relation between the Atom and the Charge of 

 Electricity carried by it. By J. J.Thomson, M.A., F.R.S., 

 Professor of Experimental Physics, Cambridge*. 



IN the electrolysis of solutions, the persistency of the sign 

 of the electric charge carried by an ion is almost as 

 marked a feature as the constancy of the magnitude of the 

 charge. Thus the hydrogen ion always carries a positive 

 charge, the chlorine ion a negative one. In the electrolysis 

 of gases, however, the sign of the charges carried by the 

 atoms of the different elements is much more variable : here 

 an atom of hydrogen does not always carry a positive charge, 

 nor an atom of chlorine always a negative one ; each of these 



* Communicated by the Author. 



2N2 



