早教吧 育儿知识 作业答案 考试题库 百科 知识分享

两个半径均为R的圆形平板电极,平行正对放置,相距为d,极板间电压为U,板间电场可以认为是均匀的.一个α粒子从正极板边缘以某一初速度垂直于电场方向射入两极板之间,到达负极板

题目详情
两个半径均为R的圆形平板电极,平行正对放置,相距为d,极板间电压为U,板间电场可以认为是均匀的.一个α粒子从正极板边缘以某一初速度垂直于电场方向射入两极板之间,到达负极板时恰好落在极板中心.已知质子电荷为e,质子和中子的质量均视为m,忽略重力和空气阻力的影响.求:
(1)极板间的电场强度E;
(2)α粒子在极板间运动的加速度a;
(3)α粒子的初速度v0



0
▼优质解答
答案和解析
(1)极间场强E=
U
d

故极板间的电场强度E=
U
d

(2)α粒子在极板间运动的加速度a=
qE
m
,代入得:
       a=
eU
2md

故α粒子在极板间运动的加速度a=
eU
2md

(3)由d=
1
2
at2得:t=
2d
a
=2d
m
eU
v0=
R
t
R
2d
eU
m

故α粒子的初速度v0=
R
2d
eU
m
E=
U
d
UUUddd
故极板间的电场强度E=
U
d

(2)α粒子在极板间运动的加速度a=
qE
m
,代入得:
       a=
eU
2md

故α粒子在极板间运动的加速度a=
eU
2md

(3)由d=
1
2
at2得:t=
2d
a
=2d
m
eU
v0=
R
t
R
2d
eU
m

故α粒子的初速度v0=
R
2d
eU
m
E=
U
d
UUUddd
(2)α粒子在极板间运动的加速度a=
qE
m
,代入得:
       a=
eU
2md

故α粒子在极板间运动的加速度a=
eU
2md

(3)由d=
1
2
at2得:t=
2d
a
=2d
m
eU
v0=
R
t
R
2d
eU
m

故α粒子的初速度v0=
R
2d
eU
m
qE
m
qEqEqEmmm,代入得:
       a=
eU
2md

故α粒子在极板间运动的加速度a=
eU
2md

(3)由d=
1
2
at2得:t=
2d
a
=2d
m
eU
v0=
R
t
R
2d
eU
m

故α粒子的初速度v0=
R
2d
eU
m
a=
eU
2md
eUeUeU2md2md2md
故α粒子在极板间运动的加速度a=
eU
2md

(3)由d=
1
2
at2得:t=
2d
a
=2d
m
eU
v0=
R
t
R
2d
eU
m

故α粒子的初速度v0=
R
2d
eU
m
eU
2md
eUeUeU2md2md2md
(3)由d=
1
2
at2得:t=
2d
a
=2d
m
eU
v0=
R
t
R
2d
eU
m

故α粒子的初速度v0=
R
2d
eU
m
d=
1
2
111222at2得:t=
2d
a
=2d
m
eU
v0=
R
t
R
2d
eU
m

故α粒子的初速度v0=
R
2d
eU
m
2得:t=
2d
a
=2d
m
eU
v0=
R
t
R
2d
eU
m

故α粒子的初速度v0=
R
2d
eU
m
t=
2d
a
2d
a
2d
a
2d
a
2d2d2daaa=2d
m
eU
m
eU
m
eU
m
eU
mmmeUeUeUv0=
R
t
R
2d
eU
m

故α粒子的初速度v0=
R
2d
eU
m
v0=
R
t
R
2d
eU
m

故α粒子的初速度v0=
R
2d
eU
m
0=
R
t
RRRttt=
R
2d
RRR2d2d2d
eU
m
eU
m
eU
m
eU
m
eUeUeUmmm
故α粒子的初速度v00=
R
2d
eU
m
R
2d
RRR2d2d2d
eU
m
eU
m
eU
m
eU
m
eU
m
eUeUeUmmm.
看了两个半径均为R的圆形平板电极,...的网友还看了以下: