In the figure, a uniform, upward-pointing electric field E of magnitude 4.50×103
ID: 2046841 • Letter: I
Question
In the figure, a uniform, upward-pointing electric field E of magnitude 4.50×103 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 = 4 cm and separation d = 2.00 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity v0 of the electron makes an angle ?=45° with the lower plate and has a magnitude of 9.17×106 m/s.The next electron has an initial velocity which has the same angle ?=45° with the lower plate and has a magnitude of 6.32×106 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 ( m).
Explanation / Answer
E = 4.50 × 10^3 N/C, L = 4 cm, d = 2.00 cm. The first electron has the initial velocity v = 9.17 × 10^6 m/s, A = 45°, Will this electron strike one of the plates? vertical acceleration a = F/m = qE/m = 4.4*10^14 m/s^2 maximum height h = [v*sin(A)]^2/(2a) = 2.32 cm > d so it will strike the upper plate. when it strikes the upper plate, time = t vertical displacement = d = vsin(A)t - at^2/2 0.02 = 4.51*10^6*t - 2.2*10^14 t^2 2.2*10^14 t^2 - 4.51*10^6*t + 0.02 = 0 t = 6.49*10^-9 s horizontal displacement = vcos(A)*t = 2.93 cm which is the horizontal distance from the left edge.