In the figure below, a uniform, upward electric field E rightarrow of magnitude
ID: 2127329 • Letter: I
Question
In the figure below, a uniform, upward electric field E rightarrow of magnitude 2.00 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 = 11 cm and separation d = 1.9 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity v rightarrow 0 of the electron makes an angle ? = 57 degree with the lower plate and has a magnitude of 6.10 times 106 m/s With M = 9.1 times 10^-31 (mass of an electron) Q = 1.601 times 10^-19 (charge of an electron) E = 2.00 times 10^3 (given)Explanation / Answer
The electron has a uniform
motion along x-axis and a
decelerate motion along y-axis
because of the net force C mg F
r r
+ .
Along the x-axis: vox = vo cosq x voxt =
Along the y-axis: sinq oy o v = v 2
2
1
y voyt at = ? and v v at y oy = ?
The accelerated motion along y-axis is due to a net force: ma = eE + mg
( )( ) 2 14 2
31
19
9.81 / 3.51 10 /
9.11 10
1.6 10 2 10 /
m s m s
kg
C N C
E g
m
e
a + = *
*
* *
= + =
?
?
Y
q
L
O xmax X
d
Fc
mg
v0
E
At y = ymax , = 0 y v and from equation oy y v ? at = v , we get vo at sinq =
the time the charge needs to travel upward is
a
v
t o sinq
= .
We introduce this time in the equation of motion: 2
max
2
1
y voyt at = ?
And solve for max y
a
v
a
v
a
a
v
y o o o
2
sin
2
sin sin 2 2
2
2 2 2 2
max
q q q
= ? =
So,
( )
( )
m
m s
m s
y 2
14 2
6 2
max 2.56 10
4 3.51 10 /
6 10 / ? = *
*
*
= .
Because y 2.56 cm d 2.0 cm max = > = , it means that the electron strikes the
upper plate, first !!