In the figure, a uniform, upward-pointing electric field E of magnitude 4.50×10
ID: 1406934 • Letter: I
Question
In the figure, a uniform, upward-pointing electric field E of magnitude 4.50×103 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 9.51×106 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.
Another electron has an initial velocity which has the angle =45° with the lower plate and has a magnitude of 6.44×106 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
L = 4 cm = 0.04 m ; d = 2.00 cm = 0.02 m, E = 4.5 x 103 N/s
Case 1
vo = 9.51 x 106 m/s; =45°
Horizontal component of velocity of electron = vox = vo cos 45 = 9.51 x 106 cos 45 = 6.7 x 106 m/s
Vertical component of velocity of electron = voy = vo sin 45 = 9.51 x 106 sin 45 = 6.7 x 106 m/s
Time taken by electron to cover vertical distance d = 0.02 m = ty
where y = voyty +0.5 aty2 = 6.7x106ty + 0.5 (qE/m) ty2 [ F= ma; qE = ma ; a = qE/m]
0.02 = 6.7 x 106ty+ 0.5(1.6 x 10-19x 4.5 x 10/9.1 x 10-31) ty2
0.02 = 6.7 x 106ty + 0.3956 x 10^15ty2
0.3956 x 1015ty2 + 6.7 x 106ty - 0.02 = 0
ty = 2.589 x 10-9 s
Time taken by electron to cover horizontal distance x = L = 0.04 cm is tx = ?
tx = horizontal distance/ horizontal component of velocity = 0.04/6.7 x 10^6 = 5.97 x 10^-9 s
Since time taken by electron to cover vertical distance is less than time taken to cover horizontal distance, it will strike the plate.
Let horizontal distance from the left edge = x' = voxty = 6.7 x 106 x 2.589 x 10-9 s = 0.0173 m = 1.73 cm from left edge
Case 2
vo = 6.44 x 106 m/s; =45°
Horizontal component of velocity of electron = vox = vo cos 45 = 6.44 x 106 cos 45 = 4.55 x 106 m/s
Vertical component of velocity of electron = voy = vo sin 45 = 6.44 x 106 cos 45 = 4.55 x 106 m/s
Time taken by electron to cover vertical distance d = 0.02 m = ty
where y = voyty +0.5 aty2 = 4.55x106ty + 0.5 (qE/m) ty2 [ F= ma; qE = ma ; a = qE/m]
0.02 = 4.55 x 106ty+ 0.5(1.6 x 10-19x 4.5 x 103/9.1 x 10-31) ty2
0.02 = 4.55x 106ty + 0.3956 x 1015ty2
0.3956 x 1015ty2 + 4.55 x 106ty - 0.02 = 0
ty = 3.394 x 10-9 s
Time taken by electron to cover horizontal distance x = L = 0.04 cm is tx = ?
tx = horizontal distance/ horizontal component of velocity = 0.04/4.55 x 10^6 = 8.79 x 10^-9 s
Since time taken by electron to cover vertical distance is less than time taken to cover horizontal distance, it will strike the plate.
Let horizontal distance from the left edge = x' = voxty =4.55x 106 x 3.394 x 10-9 s = 0.0154 m = 1.54 cm from left edge