题目内容
(14分)一质量为m、带电荷量为+q的小球以水平初速度v0进入竖直向上的匀强电场中,如图12甲所示.今测得小球进入电场后在竖直方向下降的高度y与水平方向的位移x之间的关系如图12乙所示.根据图乙给出的信息,(重力加速度为g)求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241414541021931.png)
(1)匀强电场场强的大小;
(2)小球从进入匀强电场到下降h高度的过程中,电场力做的功;
(3)小球在h高度处的动能.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241414541021931.png)
(1)匀强电场场强的大小;
(2)小球从进入匀强电场到下降h高度的过程中,电场力做的功;
(3)小球在h高度处的动能.
(1)
-
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454289985.png)
(3)
+![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454460591.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454164528.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454226816.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454289985.png)
(3)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454351778.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454460591.png)
(1)小球进入电场后,水平方向做匀速直线运动,设经过时间t,水平方向:
v0t=L
竖直方向:
=h
所以E=
-
.
(2)电场力做功为
W=-qEh=
.
(3)根据动能定理mgh-qEh=Ek-![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454460591.png)
得Ek=
+
.
v0t=L
竖直方向:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454538900.png)
所以E=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454164528.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454226816.png)
(2)电场力做功为
W=-qEh=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454289985.png)
(3)根据动能定理mgh-qEh=Ek-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454460591.png)
得Ek=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454351778.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141454460591.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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