In the figure, a uniform, upward electric field E vector magnitude 3.50 times 10
ID: 3279034 • Letter: I
Question
In the figure, a uniform, upward electric field E vector magnitude 3.50 times 10^3 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 electro is then shot between plates from the left edge of the lower plate. The initial velocity v vector_0 of the electron makes an angle theta = 45 degree with the lower plate and has a magnitude of 7.62 times 10^6 m/s. will the 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. Tries 2/10 Previous Tries The next electron has an initial velocity which has the same angle theta = 45 degree with the lower plate and has a magnitude of 5.27 times 10^6 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. Tries 0/10Explanation / Answer
Given,
E = 3.5 x 10^3 N/C ; L = 4 cm ; d = 2 cm ; theta = 45 deg ;
v0 = 7.62 x 10^6 m/s
a)The net force acting is:
Fnet = ma = e E + mg
a = e E/m + g
a = 1.6 x 10^-19 x 3.5 x 10^3/9.1 x 10^-31 + 9.8 = 6.15 x 10^14 m/s^2
we know that, at highest point, vy = 0, from eqn of motion
v = u + at => vy = v0y - at
t = v0y/a = v0 sin(theta)/a
we know that, y = u t + 1/2 at^2
y(max) = v0y t - 1/2 at^2
putting the value of t we get
y(max) = v0 sin(theta)/a - a v0^2 sin^2(theta)/2 a^2 = v0^2 sin^(theta)/2a
y(max) = (7.62 x 10^6)^2 sin^2(45)/2 x 6.15 x 10^14 = 0.0236 m
y(max) = 0.0236 m = 2.36 cm > d = 2 cm,
Hence, the electron strikes the upper plates.
b)for horizontal distance
d = voy t - 1/2 at^2
t^2 - 2 vo sin(theta)/a t + 2 d/a = 0
t^2 - (2 x 5.27 x 10^6 x sin45/6.15 x 10^14)t + 2 x 0.02/6.15 x 10^14 = 0
t^2 - 1.21 x 10^-8 t + 6.5 x 10^-17 = 0
t = 3.47 x 10^-9 s ; 1.77 x 10^-8
D = vox t
D = 5.27 x 10^6 x cos45 x 3.47 x 10^-9 = 0.013 m
Hence, D = 0.013 m = 1.3 cm.