题目内容
如图所示,带电平行金属板A、B,板间的电势差为U,A板带正电,B板中央有一小孔.一带正电的微粒,带电量为q,质量为m,自孔的正上方距板高h处自由落下,若微粒恰能落至A、B两板的正中央c点,不计空气阻力,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241552544692242.png)
则:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241552544692242.png)
则:
A.微粒在下落过程中动能逐渐增加,重力势能逐渐减小 |
B.微粒在下落过程中重力做功为![]() ![]() |
C.微粒落入电场中,电势能逐渐增大,其增加量为![]() |
D.若微粒从距B板高1.5h处自由下落,则恰好能达到A板 |
C
试题分析:微粒在下落过程中先做加速运动,后做减速运动,动能先增大,后减小.重力一直做正功,重力一直减小.故A错误.
微粒下降的高度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155254844569.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241552549371029.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155255187929.png)
微粒落入电场中,电场力做负功,电势能逐渐增大,其增加量等于微粒克服电场力做功
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155255281740.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241552553741114.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241552554681177.png)
点评:题根据动能定理研究微粒能否到达A板,也可以根据牛顿第二定律和运动学公式分析.
![](http://thumb2018.1010pic.com/images/loading.gif)
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