Question

Question 3 of 3 A 0.0380 g positively charged ball with charge q = 4.15 μC is resting on a flat, frictionless, horizontal surface. For a time of t = 0.0380 s, a constant electric field of magnitude E = 795 N / C is directed vertical to the ball, which makes the ball rise to a height of d . After this time, the electric field is turned off, and the ball returns to the surface. Find the height in meters by which the ball is able to be lifted off the surface.

Answer 1

First the acceleration will be due to both the gravity and electric field. Once the electric field is turned off, only acceleration due to gravity will act.

Ang mass of charge, m = 0.0380 g = 3.80 x 1050kg charge, q = 4.15 uC . 1 . = 4.15 410oC. Electric field, t = 795 N/C Force on the charge due to electric field, & E=qE Also -=ma Newton's I 1 } a law a= acceleration ubwards where A e-mg = ma ma = ma 9 E-mg - ma 4.15*166 x 795 - 9,8 = a 3.80 x 100 a = 77.02 m/s² Electric field is turned off t = 0.0 3 8 8 after TV Max height t=0.0385 - Vozo

Height attained by the particle in time Dot = 0.03803 h = Vort &t at 20+ x (77.02)x(0.038) _ 0.0556 m Now, electic field is turned off. Gravity will act d at this moment v = vtat O + 77.02 X 0.038 = Qeparois mois = 2.92676 mis Y = y2 - 29 ha o= (2.92 676)2 - 2x9.8 h h = 0.437 m & Total height the ball is lifted = uitha = 0.0556 +0.437 = 0,4926 m = 0.49 m Answer