Question

[ Course Contents » ... » HW2 » An electron is then shot between the plates Notes Bookmark Evaluate Communicate Print Info FA POR In the figure, a uniform, upward-pointing electric field E of magnitude 5.00x103 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 = 4 cm and separation d = 2.00 cm. Electrons are shot between the plates from the left edge of the lower plate. The first electron has the initial velocity Vo, which makes an angle 0=45° with the lower plate and has a magnitude of 9.92x106 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates. Submit Answer Tries 0/10 Another electron has an initial velocity which has the angle 0=45° with the lower plate and has a magnitude of 6.29x106 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates. Submit Answer Tries 0/10

Answer 1

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E = 5x133 / d=2cm L= 4cm a) v = 9.92x106m sec Ja 2cm = 1.6x109 x 5x103 9.1x163) = 8.79123104m seca 4cm. (R) = horizontal = Vo sinas - (9.92816) ² distance - = 0.112m 8.79128, =11.2cm - Equation of trajectory, of projectile, h = (tano) x - x ² 207600? h = (ten 43°) x - 8.7912 x10!4xx? 3x19.92X106)?coms h = x -8.93356x² for relea, the h=zem = 0.02m. 0.02 = 2-8.93356x? x =0.026 m and 0.08586 m [x = 2.6em and 8.586cm yes electron will hit the upper plate) at horizontal distance x=2.6cm. Amwhere

6) Vo = 6-2 9X106 m Nowe, Prague Equation of trajectory, h = fan45x- ax² , h=x- 8.79/2810142 -=x-22.22x² 76.29X10692 for h=2cm = 0.02=X-22.22x² x=) no real values. Means election will not hit any plates. & at u=4 cm, h= 0.04 - 22.22 (0.04) 2 h - 4.448x10²m th=0.4448 cm