When the electric field was applied, the rays were deflected and a part of the pencil of rays was stopped by the mica screen. A shadow was thus cast on the plate which showed the direction of deviation and corresponded to the least deviable rays which gave an impression through the black paper. If a particle of mass _m_, charge _e_, and velocity _u_, is projected normally to an electric field of strength _X_, the acceleration α is in the direction of the field, and is given by _Xe_ α = ----- . _m_ Since the particle moves with a constant acceleration parallel to the field, the path of the particle is the same as that of a body projected horizontally from a height with a constant velocity and acted on by gravity. The path of the particle is thus a parabola, whose axis is parallel to the field and whose apex is at the point where the particle enters the electric field. The linear deviation _d₁_ of the ray parallel to the field after traversing a distance _l_ is given by 1 _Xe_ _l²_ _d₁_ = -- -- ---- ---- . 2 _m_ _u²_ On leaving the electric field, the particle travels in the direction of the tangent to the path at that point. If θ is the angular deviation of the path at that point _eXl_ tan θ = ----- . _mu²_ The photographic plate was at a distance _h_ above the extremity of the field. Thus the particles struck the plate at a distance _d₂_ from the original path given by $$ d_2 = h \tan \theta + d_1 $$ $$ = \frac {Xle} {mu^2} (\frac {l} {2} + h) $$ In the experimental arrangement the values were _d₂_ = ·4 cms.; _X_ = 1·02 × 10¹²; _l_ = 3·45 cms.; _h_ = 1·2 cms. If the radius _R_ of curvature of the path of the same rays is observed in a magnetic field of strength _H_ perpendicular to the rays, _e_ _V_ --- = ---- _m_ _HR_ Combining these two equations we get $$ u = \frac {X . l (\frac {l} {2} + h)} {H . R . d_2} $$ . A difficulty arose in identifying the part of the complex pencil of rays for which the electric and magnetic deviations were determined. Becquerel estimated that the value of _HR_ for the rays deflected by the electric field was about 1600 C.G.S. units. Thus _u_ = 1·6 × 10¹⁰ cms. per second, and _e_ ---- = 10⁷. _m_