解题思路:
(1)微粒在加速电场中运动由动能定理得:
q
U
1
=
mv
①解得
v
0
=
1.0
×
10
4
m
/
s
(2)微粒在偏转电场中做类平抛运动,有:
a
=
,
v
y
=
a
t
=
a
飞出电场时,速度偏转角的正切为:
tan
θ
=
=
=
=
②解得
θ
=
30
∘
1.0×10 4m/s 30°
<>
解题思路:
(1)微粒在加速电场中运动由动能定理得:
q
U
1
=
mv
①解得
v
0
=
1.0
×
10
4
m
/
s
(2)微粒在偏转电场中做类平抛运动,有:
a
=
,
v
y
=
a
t
=
a
飞出电场时,速度偏转角的正切为:
tan
θ
=
=
=
=
②解得
θ
=
30
∘
1.0×10 4m/s 30°
<>