题目内容
如图所示,板长为L的平行板电容器倾斜固定放置,极板与水平线夹角θ=30°,某时刻一质量为m、带电荷量为q的小球由正中央A点静止释放,小球离开电场时速度是水平的(提示:离开的位置不一定是极板边缘),落到距离A点高度为h的水平面处的B点,B点放置一绝缘弹性平板M,当平板与水平夹角α=45°时,小球恰好沿原路返回A点.求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250036115533340.jpg)
(1)电容器极板间的电场强度E;
(2)平行板电容器的板长L;
(3)小球在A、B间运动的周期T.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250036115533340.jpg)
(1)电容器极板间的电场强度E;
(2)平行板电容器的板长L;
(3)小球在A、B间运动的周期T.
(1)
(2)3h (3)2![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250036115991097.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611568758.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250036115991097.png)
(1)带电粒子沿水平方向做匀加速运动,可知:
qEcos θ=mg ①
故E=
.
(2)小球垂直落到弹性挡板上,且α=45°
则有v0=vy=
②
由动能定理得:qE·
Ltan θ=
mv
③
由②③得:L=3h.
(3)由小球在电场中做匀加速运动
=
gtan θ·t![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611740281.png)
t1=
=![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611849597.png)
平抛运动的时间为t2=![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611865587.png)
总时间为:
t=2t1+2t2=2
.
qEcos θ=mg ①
故E=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611568758.png)
(2)小球垂直落到弹性挡板上,且α=45°
则有v0=vy=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611631533.png)
由动能定理得:qE·
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611646338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611646338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611677293.png)
由②③得:L=3h.
(3)由小球在电场中做匀加速运动
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611693596.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611646338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611740281.png)
t1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611833727.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611849597.png)
平抛运动的时间为t2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003611865587.png)
总时间为:
t=2t1+2t2=2
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250036115991097.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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