题目内容
(20分)匀强电场的方向沿x轴正向,电场强度E随x的分布如图所示。图中E0和d均为已知量。将带正电的质点A在O点由能止释放。A离开电场足够远后,再将另一带正电的质点B放在O点也由静止释放,当B在电场中运动时,A、B间的相互作用力及相互作用能均为零;B离开电场后,A、B间的相作用视为静电作用。已知A的电荷量为Q,A和B的质量分别为m和
。不计重力。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241343083554301.png)
(1)求A在电场中的运动时间t,
(2)若B的电荷量q =
Q,求两质点相互作用能的最大值Epm
(3)为使B离开电场后不改变运动方向,求B所带电荷量的最大值qm
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308293319.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241343083554301.png)
(1)求A在电场中的运动时间t,
(2)若B的电荷量q =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308496302.png)
(3)为使B离开电场后不改变运动方向,求B所带电荷量的最大值qm
(1)
(2)
QE0d (3)
Q
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308574559.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308496302.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308745330.png)
解:(1)由牛顿第二定律得,A在电场中的加速度 a =
= ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309026409.png)
A在电场中做匀变速直线运动,由d =
a
得
运动时间 t =
= ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308574559.png)
(2)设A、B离开电场时的速度分别为vA0、vB0,由动能定理得
QE0d =
m![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309369400.png)
qE0d =![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309541521.png)
A、B相互作用过程中,动量和能量守恒。A、B相互作用为斥力,A受力与其运动方向相同,B受的力与其运动方向相反,相互作用力对A做正功,对B做负功。A、B靠近的过程中,B的路程大于A的路程,由于作用力大小相等,作用力对B做功的绝对值大于对A做功的绝对值,因此相互作用力做功之和为负,相互作用能增加。所以,当A、B最接近时相互作用能最大,此时两者速度相同,设为v,,
由动量守恒定律得:(m +
)v, = mvA0 +
vB0
由能量守恒定律得:EPm = (
m
+
)—
)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134310430325.png)
且 q =
Q
解得相互作用能的最大值 EPm =
QE0d
(3)A、B在x>d区间的运动,在初始状态和末态均无相互作用
根据动量守恒定律得:mvA +
vB = mvA0 +
vB0
根据能量守恒定律得:
m
+
=
m
+ ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309541521.png)
解得:vB = -
+ ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134311117438.png)
因为B不改变运动方向,所以vB = -
+
≥0
解得: q≤
Q
则B所带电荷量的最大值为:qm =
Q
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308979312.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309026409.png)
A在电场中做匀变速直线运动,由d =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309073284.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309167287.png)
运动时间 t =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309213451.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308574559.png)
(2)设A、B离开电场时的速度分别为vA0、vB0,由动能定理得
QE0d =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309073284.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309369400.png)
qE0d =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309541521.png)
A、B相互作用过程中,动量和能量守恒。A、B相互作用为斥力,A受力与其运动方向相同,B受的力与其运动方向相反,相互作用力对A做正功,对B做负功。A、B靠近的过程中,B的路程大于A的路程,由于作用力大小相等,作用力对B做功的绝对值大于对A做功的绝对值,因此相互作用力做功之和为负,相互作用能增加。所以,当A、B最接近时相互作用能最大,此时两者速度相同,设为v,,
由动量守恒定律得:(m +
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308293319.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308293319.png)
由能量守恒定律得:EPm = (
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309073284.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309369400.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309541521.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134310383512.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134310430325.png)
且 q =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308496302.png)
解得相互作用能的最大值 EPm =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308496302.png)
(3)A、B在x>d区间的运动,在初始状态和末态均无相互作用
根据动量守恒定律得:mvA +
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308293319.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308293319.png)
根据能量守恒定律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309073284.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134310695383.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134310758500.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309073284.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309369400.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134309541521.png)
解得:vB = -
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134311039420.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134311117438.png)
因为B不改变运动方向,所以vB = -
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134311039420.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134311117438.png)
解得: q≤
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308745330.png)
则B所带电荷量的最大值为:qm =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134308745330.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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