Scattering of Positive Electricity by Light. 441 



the light, as the deflexion of the electrometer on the scale 

 could not then be read off. Therefore the plate was exposed 

 only five seconds, and the deflexion — reduced to volts — was 

 multiplied by 12*. But beside this action of the light — 

 undoubted and already known — the numbers show no other. 

 There are, it is true, deflexions of the electrometer-needle in 

 the dark as well as in the light, which, however, in no case 

 reach the amount of one volt, and which on account of their 

 inconstancy are to be referred to an irregular passage of 

 electricity from the gauze to the plate, probably caused by 

 the dust of the air. 



Only in two cases is this feeble transference of electricity 

 slightly greater in light than in the dark, viz. with a nega- 

 tively charged gauze opposed to a plate of amalgamated zinc, 

 and with the gauze positive and the plate of paraffined zinc. 

 If one wishes to find in this a proof of an action of light, 

 then only the first case can be taken to show a photoelectric 

 dispersion of positive electricity. But here also a sufficient 

 explanation is to be found in the fact that the ultra-violet 

 light reflected from the polished surface of the amalgamated 

 zinc strikes the side of the gauze turned towards it and pro- 

 duces a passage of negative electricity from it to the plate, 

 so that in this case the photoelectric discharge is not from the 

 positively charged plate but from the negatively charged 

 gauze. 



We have now to describe an experimental arrangement in 

 which this action of the reflected light appears perfectly 

 clearly. All the observations show that the paraffined or 

 greased surfaces are not photoelectrically sensitive ; in no 

 case is the scattering of electricity from these found to be 

 greater with a positive charge than with a negative charge 

 or in the dark. 



The small deflexions of the electrometer observed in the 



* A more accurate calculation of the change of potential during an ex- 

 posure of one minute would be obtained by use of the formula 



hit. 



V- Vl =V.e- 



in which V denotes the potential of the charging battery, V, that of the 

 illuminated plate, k a constant, J the intensity of the light, and t x the 

 time of exposure. From this we should have for two times of exposure 

 t x and t 2 and the corresponding potentials v x and v 2 : 



M^HM^} 



from which v 2 can be easily calculated. In the foregoing case we obtain 

 for v 2 the value shown in brackets, -J- 123 volts. 



