Question

An electron is projected as in fig. 3-10 at a speed of 6.0106 m/s at an angle  of 45o ; E = 2.0103 N/C (directed upward), d = 2.0 cm, and l = 10.0 cm. (a) Will the electron strike either of the plates? (b) If it strikes a plate, where does it do so? 

Answer 1

Acceleration of the electron will be:

$a = \frac{eE}{m} =3.516 \times 10^{14}\;m/s^2$

This acceleration will be in the downward direction.

For horizontal direction:

$t = \frac{l}{v_0cos45}=\frac{0.1}{6 \times 10^6cos45}=2.357 \times 10^{-8} \;s$

The vertical distance traveled by the electron will be:

$H = u_yt-\frac{1}{2}at^2 \\ = (6 \times 10^6sin45) \times 2.357 \times 10^{-8}+\frac{1}{2}(3.516 \times 10^{14}) \times (2.357 \times 10^{-8})^2=0.1976 \;m = 19.76 \;cm$

which is much greater then d = 2 cm

Hence the electron will strike the plate.

Substitute H=d=0.02 m

$0.02=(6 \times 10^6sin45)t+ \frac{1}{2}(2.516 \times 10^{14})t^2$

Solve this quadratic equation to get:

$t = 4.0383 \times 10^{-9}\;s \\ x=6 \times 10^6cos45 \times 4.0383 \times 10^{-9}=0.01713 \;m=1.713 \;cm$

So, the electron will strike 1.713 cm from the left.