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()
2
2
66
1
3
2
22
00
2
14
9
(4.24 10 m s)
4.24 10 m s
1.40 10 m s
sin
sin
2
3.51 10 m s
6.43 10 s.
vv
a
d
t
a
θθ
−
×−
×
−
×
−−
==
×
=×
The negative root was used because we want the
earliest
time for which
y
=
d
. The
x
coordinate is
( )
69
2
0
cos
6.00 10 m s 6.43 10 s cos45
2.72 10 m.
xv
t
×
×
°
=
×
θ
This is less than
L
so the electron hits the upper plate at
x
= 2.72 cm.
86. (a) The electric field is upward in the diagram and the charge is negative, so the force
of the field on it is downward. The magnitude of the acceleration is
a
=
eE
/
m
, where
E
is
the magnitude of the field and
m
is the mass of the electron. Its numerical value is
a
=
××
×
−
−
160 10
911 10
351 10
19
31
14
.
.
..
C2
.
0
01
0NC
kg
ms
3
2
ch
c
h
We put the origin of a coordinate system at the initial position of the electron. We take
the
x
axis to be horizontal and positive to the right; take the
y
axis to be vertical and
positive toward the top of the page. The kinematic equations are
2
0
1
cos ,
sin
,
and
sin
.
2
y
x
vt
y vt
a
t
v
v
a
t
−
=
−
First, we find the greatest
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Charge, Acceleration, Force, Mass

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