In the figure, a uniform, upward-pointing electric field E of magnitude 3.00x103
ID: 1880373 • Letter: I
Question
In the figure, a uniform, upward-pointing electric field E of magnitude 3.00x103 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 vo, which makes an angle -450 with the lower plate and has a magnitude of 7.46x106 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. Submit Answer Tries 0/10 Another electron has an initial velocity which has the angle =45° with the lower plate and has a magnitude of 5.92×106 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. Submit Answer Tries 0/10Explanation / Answer
a) let;s start with finding retardation for electron
F = qE = ma
a= qE/ m = -1.6 x 10^-19 (3000)/ ( 9.109 x 10^-31)= 5.27 x 10^ 14 m/s ^2
Uy = 7.46 x 10^6 sin 45 = 5.275 x 10^6 m/s
V^2 = u^2 - 2ah
v^2= ( 5.275 x 10^6)^2 - 2 (5.27 x 10^ 14 ) (2/100) = 27.825 x 10^12 - 21.08 x 10^12 which will give a positive value indicating that electron will touch the negative plate.
velocity with which electrion will touch the plate= 2.6 x 10^6 m/s
time= ( v-U)/a= ( 2.6 x 10^6- 5.275 x 10^6)/ (5.27 x 10^ 14) = 0.508 x 10^ -8 seconds apprx
horizontal distance= v t = 5.275 x 10^6 m/s ( 0.508 x 10^ -8)=2.7 cm apprx from the left edge of the lower plate
b)initial vertical velocity = 5.92 x 10^6 sin 45= 4.186 x 10^ 6 m/s
retardation= - 5.27 x 10^ 14 m/s ^2
let's work out if the electron will touich the upper plate,
v^2 = 4.186^2 x 10^12 - 2 (5.27 x 10^ 14 ( 2/100) which gives a negative value therefore electron in this case will mnot be able to styrike the plate.
let's work out the position
v^2 = u^2 - 2ah
0 = 4.186^2 x 10^12 - 2 ((5.27 x 10^ 14) h
h = 1.66 cm from the lower plate