题目内容
如图所示,两平行金属板A、B水平放置,两板间的距离d=40cm.电源电动势E=24V,内电阻r=1Ω,电阻R=15Ω.闭合开关S,待电路稳定后,在两板之间形成匀强电场.在A板上有一个小孔k,一个带电荷量为
C、质量为
kg的粒子P由A板上方高h=10cm处的O点自由下落,从k孔进入电场并打在B板上
点处.当P粒子进入电场时,另一个与P相同的粒子Q恰好从两板正中央
点水平飞入.那么,滑动变阻器接入电路的阻值为多大时,粒子Q与P恰好同时打在
处。 (粒子间的作用力及空气阻力均忽略不计,取g=10m/s2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250038262903590.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003826212577.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003826228619.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003826259350.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003826259333.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003826259350.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250038262903590.jpg)
滑动变阻器接入电路的阻值为8Ω
试题分析:根据机械能守恒定律可得P粒子进入电场时的速度为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003826306995.png)
设P、Q在电场中运动的加速度为a,运动到打在
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003826259350.png)
对P:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003826337807.png)
对Q:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003826352724.png)
联立解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003826368559.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003826384663.png)
而
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003826399822.png)
所以,滑动变阻器两端的电压为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250038264151892.png)
由欧姆定律可得通过滑动变阻器的电流为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250038264301161.png)
所以,滑动变阻器接入电路的阻值为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003826446719.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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