题目内容
(11分)如图所示,A.B为平行板电容器,两板相距d,接在电压为U的电源上,在A板的中央有一小孔M(两板间电场可视为匀强电场).今有一质量为m的带电质点,自A板上方与A板相距也为d的O点由静止自由下落,穿过小孔M后到达距B板
的N点时速度恰好为零.(重力加速度为g)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250013528675661.jpg)
求:(1)带电质点的电荷量,并指出其带电性质;
(2)在保持与电源相连的情况下,A板往下移
的距离.质点仍从O点由静止自由下落,求质点下落速度为零时距B板的距离.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001352852433.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250013528675661.jpg)
求:(1)带电质点的电荷量,并指出其带电性质;
(2)在保持与电源相连的情况下,A板往下移
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001352883410.png)
(1)
负电 (2) ![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001352914601.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001352899803.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001352914601.png)
试题分析:(1)由题意知,质点带负电 ①
两极板间的场强E:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001352945534.png)
设电量大小为q,则从O到N点,则动能定理可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250013529611055.png)
由①②解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001352899803.png)
(2)当A板下移
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001352883410.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001353008703.png)
两板间的场强
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001353039346.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001353055569.png)
设下落速度为零时距B板距离为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001353070426.png)
则从开始下落到速度为零的过程中,由动能定理得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250013530861387.png)
由④⑤⑥⑦得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001352914601.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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