In the figure, a uniform, upward-pointing electric field E of magnitude 4.00×10

Question

In the figure, a uniform, upward-pointing electric field E of magnitude 4.00×103 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. Electrons are shot between the plates from the left edge of the lower plate.  
The first electron has the initial velocity v0, which makes an angle ?=45° with the lower plate and has a magnitude of 8.74×106 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.
2.64×10-2 m

Another electron has an initial velocity which has the angle ?=45° with the lower plate and has a magnitude of 6.18×106 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.

Explanation / Answer

Given that

The upward-pointing electric field E of magnitude (E) = 4.00×103 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 =0.04m

The distance between the plates is (d) = 2.00 cm =0.02m

Here when a electron is projected with a velocity vo and angle 45 degrees with the horizontal plate, therefore the velocity is resolved into two componwnts

vx =vocostheta =8.74×106 m/s*cos45=6.180*106m/s

vy =vosintheta=8.74×106 m/s*cos45 =6.180*106m/s

The vertical acceleration of the particle to reach the other plate is given by

F =qE force acting on an electron in an electric field

ma=qE then acceleration a =qE/m =(1.6*10-19)(4.00×103 N/C )/9.1*10-31kg=0.703*1015m/s2 =7.03*1014m/s2

Now to find the time taken for the electron to reach the opposite plates and on the lower plate to reach the other end is given by

sy =vyty+(1/2)aty2====> 0.02 =(6.180*106m/s)ty+(0.5)(7.03*1014m/s2)ty2 ===>0.02 =(6.180*106m/s)ty+(3.515*1014m/s2)ty2

Solving this quadratic equation we get time taken for the electron to reach the opposite plate is (ty) =0.0000305305s =3.05*10-9s

Now the horizontal time period for the electron to travel on the lower plate is given by tx is from

sx =vxtx+(1/2)atx2====>0.04 =(6.180*106m/s)tx ====> tx= 6.47*10-9s

From the above two time period values ty<tx, the electron will strike the horizontal plate x =(6.180*106m/s)(3.05*10-9s)=0.018m.

similalarly repat the second for finding the values you get the required answer.

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