Question

In the figure a uniform, upward electric field of magnitude 1.80 × 10³ N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 13.0 cm and separation d = 1.70 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity of the electron makes an angle θ = 46.0° with the lower plate and has a magnitude of 6.50 × 10⁶ m/s. (a) Will the electron strike one of the plates? (b) If so, which plate, top or bottom? (c) How far horizontally from the left edge will the electron strike?

Answer 1

E = 1.80 x10NC v - 6.5x109 m/s 6=46 L = 13.0 cm=0.13 m d=1.70 cm=0.017 m 9.-1.6x10C m. = 9.1x10 kg Part a The horizontal component of the initial velocity of the electron is, . y,cose = 6.5 x10 m/s) cos46 =(6.5x10 m/s) (0.694658) = 4.51x10 m/s The vertical component of the initial velocity of the electron is, vy, sine =(6.5x10 m/s sin 46 = 16.5x10 m/s)(0.71933) = 4.67 x10 m/s The net force acting on the electron due to the electric field between the two plates is, F=9E m.a=9E Here, a is the vertical acceleration of the electron Rearrange the above equation for a and substitute the values in the equation. (1.6x10*C)(1.80x10' N/C) 9.1x10- kg =-3.16x10 m/s As the electron moves up from the parallel to the direction of the electric field, the electron moves with negative acceleration or deceleration Use the following equation to find the maximum vertical displacement of the electron v-4° = 2as V-v2 = 2ay Rearrange the above equation for y- and substitute the values in the equation. 2a (0) – (4.67x10 m/s) 2(-3.16 *10* m/s) = 0.0345 m =3.45 am As the maximum vertical distance, the electron can travel is more than the separation of the plate, the electron may hit the upper plate. Therefore, the electron strikes one of the plates.

Part b The maximum vertical distance the electron can travel is, y.. =3.45 cm The separation of the two plates is, d=1.70 cm As the maximum vertical distance, the electron can travel is more than the separation of the plate, the electron may hit the upper plate. Therefore, the electron may hit upper plate. Partc Use the kinematic equation for the time taken by the electron to strike the upper plate. d=v1+ at Substitute the values in the above equation (0.017 m)= (4.67 x10 m/s)t-(3.16x104 m/s° ) (0.017) = (4.67x106)2-(1.58x1044)?? (1.58 x10*) 7 -(4.67x109)+(0.017) = 0 The solution of the above quadratic equation is, t = 4.25x10 s, 2.53x10 s Here, the time 4.25x10 s is taken to strike the upper plate and 2.53x10 s is the time taken by the electron while coming down at d, if there is upper plate. The horizontal distance travelled by the electron is, X=V. =(4.51x109 m/s)(4.25x10*s) = 0.01916 m 100.com =1.916 cm Therefore, the electron strikes the upper plate at a horizontal distance from the left edge is 1.916 cm