题目内容
如图所示,在沿水平方向的匀强电场中有一固定点O,用一根长度为L=0.40m的绝缘细线把质量为m=0.20kg,带有正电荷的金属小球悬挂在O点,电荷量q = 0.5C小球静止在B点时,细线与竖直方向的夹角为θ=37°.现将小球拉至
位置A使细线水平后由
静止释放,求:
(1)匀强电场的场强大小;
(2)小球通过最低点C时细线对小球的拉力大小.
(取g = 10m/s2,sin37°=0.60,cos37°=0.80)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241227417733548.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412274174272.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412274175872.gif)
(1)匀强电场的场强大小;
(2)小球通过最低点C时细线对小球的拉力大小.
(取g = 10m/s2,sin37°=0.60,cos37°=0.80)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241227417733548.jpg)
(1)取小球为研究对象,静止时
= tanθ ①
解得:E=3N/C
②
(2)小球从A到C的运动过程由动能定理得:
mgL –EqL =
mv2
③
到达最低点时竖直方向由牛顿第二定律得:
FT-mg = m
④
解得:FT =" 3mg-2qE" =" 3N " ⑤
评分标准:①③④每式2分,其余每式l分
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122741945429.gif)
解得:E=3N/C
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412274194585.gif)
(2)小球从A到C的运动过程由动能定理得:
mgL –EqL =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122741961225.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412274197665.gif)
到达最低点时竖直方向由牛顿第二定律得:
FT-mg = m
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122741992245.gif)
解得:FT =" 3mg-2qE" =" 3N " ⑤
评分标准:①③④每式2分,其余每式l分
略
![](http://thumb2018.1010pic.com/images/loading.gif)
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