题目内容
(11分)如图所示,在竖直放置的铅屏A的右表面上贴着
射线(即电子)放射源P,已知
射线实质为高速电子流,放射源放出
粒子的速度v0=2.0×107m/s。速度的方向沿各个方向都有,电子重量不计。足够大的荧火屏M与铅屏A平行放置,相距d=2.0×10-2m,其间有水平向左的匀强电场,电场强度大小E=1.0×105N/C。已知电子电荷量e=1.6×10-19C,电子质量取m=9.0×10-31kg。求
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241434587043030.jpg)
(1)电子到达荧光屏M上的动能为多少焦耳?
(2)电子从P点射出到达荧光屏的最长时间?
(3)荧光屏上的发光面积。(取
)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824143458424339.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824143458424339.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824143458424339.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241434587043030.jpg)
(1)电子到达荧光屏M上的动能为多少焦耳?
(2)电子从P点射出到达荧光屏的最长时间?
(3)荧光屏上的发光面积。(取
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824143458736411.png)
(1)
(2)
(3)
(2分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824143458860592.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824143458907492.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824143458985993.png)
(1)
(2分)
(3分)
(2)
(1分),
(1分) 得:
(1分)
(3)
(1分)
(2分)
本题考查带电粒子在电场中的加速问题,根据电场力做功和动能定理可求得末动能,由沿电场力方向匀加速直线运动可求得运动时间,从而求得粒子达到的半径大小
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824143459048576.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241434594691002.png)
(2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824143459516447.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824143459578323.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824143458907492.png)
(3)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824143459765953.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241434598121035.png)
本题考查带电粒子在电场中的加速问题,根据电场力做功和动能定理可求得末动能,由沿电场力方向匀加速直线运动可求得运动时间,从而求得粒子达到的半径大小
![](http://thumb2018.1010pic.com/images/loading.gif)
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