In the figure, a uniform, upward-pointing electric field E of magnitude 5.50 In
ID: 2130810 • Letter: I
Question
In the figure, a uniform, upward-pointing electric field E of magnitude 5.50
In the figure, a uniform, upward-pointing electric field E of magnitude 5.50 times 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 degree with the lower plate and has a magnitude of 1.06 times 107 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 ( m).The next electron has an initial velocity which has the same angle ?=45 degree with the lower plate and has a magnitude of 6.86 times 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
The horizontal velocity of the electron is constant and equal to 9.56 x 10^6 m/s x cos45 which is also the initial vertical component of velocity, vy. Calculate them.
Now find the vertical acceleration of the electron ay
F = Eq =ma so ay = Eq/m for the electron. Look up q,m and calculate ay. q=1.6 x 10^-19 m = 9.1 x 10^-31 I think.
Now find the time it takes the electron to reach a height equal to the top plate.
Use deltay = 0.02 m = vyt - 1/2 ay t^2. It's a quadratic, but use your calculator (I'm sure you have a quadratic solver) to find t. Take the smaller real solution.
Now find the horizontal distance travelled using deltaX = vx t If this is bigger than 0.04 m, the particle doesn't hit the top plate.
For the first one I got that the electron hits the top plate 2.70 cm in from the left. For the second, there are no roots so it never gets that high. The time then to go across the plates is 0.04/vx = 7.39 x 10^-9 s
In that time deltaY = 0.016m above the top plate.