题目内容
如图所示的示波管,电子由阴极K发射后,初速度可以忽略.经加速后水平经A板小孔,沿M、N两极板中线飞入偏转电场,最后打在荧光屏上的P点,已知加速电压为U1=2U.电源电动势E=U.内电阻r=0.R1=R,滑动变阻器的总电阻R2=R滑片位于中间位置,两偏转极板间距为L,板长为2L.从偏转极板的右端到荧光屏的距离为L,O点为两极板中线与荧光屏的交点.求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250028307285145.jpg)
(1)偏转电极MN间的电势差U2
(2)电子打在荧光屏上的偏距OP
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250028307285145.jpg)
(1)偏转电极MN间的电势差U2
(2)电子打在荧光屏上的偏距OP
(1)
;(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830759401.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830743445.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830759401.png)
试题分析:(1)由闭合电路欧姆定律:I=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830790539.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830806513.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830743445.png)
(2)在加速电场中,由动能定理:eU1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830837621.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830853703.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830868731.png)
电子在偏转电场中的运动,设偏转距离为y
在进入偏转电场的初速度方向:2L=v0t
在电场力方向:y=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830884544.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830899785.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830915434.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830946713.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830962377.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002830759401.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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