Atom and the Charge of Electricity carried hy it. 543 



greatly diminishes the potential difference between the ter- 

 minals of the induction-coil, and lowers the potential of the 

 coil surrounding the bulb to such an extent that the brush 

 discharge no longer takes place from the glass : the electric 

 intensity in the bulb is lowered, so that no further decompo- 

 sition of the gas takes place, and the bulb now does not show 

 any luminosity. The charged atoms produced by the previous 

 discharge linger for some time in the bulb before recombining, 

 and move under the electric intensity due to the oscillating 

 currents in the coil ; the lines of electric intensity arising 

 from the induction due to these currents are circles, with the 

 line joining the poles of the bulb for their axis. Under these 

 forces the atoms will tend to describe these circles ; and since 

 the negative ones move more quickly than the positive ones, 

 the negative atoms will be revolving round the axis of the 

 bulb with greater rapidity than the positive ones : hence, on 

 account of centrifugal force, there will be an excess of negative 

 atoms at the equator, arid consequently an excess of positive 

 ones at the pole : thus the equator will be negatively, the 

 poles positively electrified. This electrification is, however, 

 only transient, as the opposite charged atoms soon recombine. 



If the negative atoms move more quickly than the positive 

 ones, then in a discharge-tube in a steady state the pressure 

 at the positive electrode must be higher than that at the 

 negative. For at any part of such a tube where there is no 

 excess of positive electrification, the number of negative 

 atoms which in unit time cross a section of the tube is greater 

 than the number of positive ones. There will thus be a stream 

 of atoms towards the positive electrode : to keep the pressure 

 in the tube constant, we must compensate this by a stream of 

 molecides from the positive to the negative electrode. 



To calculate this pressure, let us suppose that the discharge- 

 tube is cylindrical of radius a. Let i be the current through 

 the tube, u the velocity of the negative ions, v the velocity of 

 the positive ; then in unit time the number of atoms lost by 

 the side of the tube next the negative electrode and gained 

 by the side of the tube next the positive is equal to 



i u — v 

 € u + v 



where e is the charge carried by one of the atoms. 



To keep the pressure constant, the positive side of the tube 

 must lose, and the negative side gain this number of mole- 

 cules. If P is the pressure-gradient, (jl the coefficient of 



