## Abstract

The determination of e by a new oil-drop method in which the electric field is horizontal has been described. The expression for e in terms of quantities measured is similar to that which applies to H. A. Wilson's method. The correction for the departure from Stokes's law is obtained from Millikan's relation e$_{1}^{\frac{2}{3}}$ = e$^{\frac{2}{3}}$ + m/pa, where e$_{1}$ is the uncorrected value of e, p cm. of mercury is the pressure and a cm. the radius of the drop. The oil drops used are larger than those used by previous experimenters, and their velocity of fall and the velocity in the direction of the electric field could be estimated with satisfactory accuracy. Assuming $\eta _{23}$ = 1830 $\times $ 10$^{-7}$ c.g.s. unit, the value of e obtained is (4$\cdot $8020 $\pm $ 0$\cdot $0013) $\times $ 10$^{-10}$ e.s.u. The probable error calculated by the least squares is about one-third of that obtained by Millikan, whose result becomes (4$\cdot $7992 $\pm $ 0$\cdot $0037) $\times $ 10$^{-10}$ e.s.u., when the above value of $\eta _{23}$ is assumed. The errors in the determination of $\eta _{23}$ by the rotating cylinder and the capillary tube method are discussed and the final mean $\eta _{23}$ = (1830$\cdot $06 $\pm $ 2$\cdot $5) $\times $ 10$^{-7}$ c.g.s. unit. is derived. This contributes an error of 1 in 500 in e and is its major uncertainty by the oil-drop method. Recent determinations of the electronic charge by the X-ray method have been analysed. The mean value of e by this method is deduced to be (4$\cdot $8044 $\pm $ 0$\cdot $0007) $\times $ 10$^{-10}$ e.s.u., and this result differs from the mean of the determinations of Millikan and the authors (4$\cdot $8007 $\pm $ 0$\cdot $002) $\times $ 10$^{-10}$ e.s.u., by 0$\cdot $0037 $\times $ 10$^{-10}$ e.s.u., which is less than the error due to the viscosity of air. Assuming the X-ray value of e, the viscosity of air can be deduced and is found to be (1830$\cdot $9 $\pm $ 0$\cdot $5) $\times $ 10$^{-7}$ c.g.s. unit at 23 degrees C.