题目内容
如右图所示,在场强E=104 N/C的水平匀强电场中,有一根长
=15 cm的细线,一端固定在O点,另一端系一个质量m=3 g、电荷量q=2×10-6 C的带正电小球,当细线处于水平位置时,小球从静止开始释放,g取10 m/s2.求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250007466261111.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250007466412884.jpg)
(1)小球到达最低点B的过程中重力势能、电势能分别变化了多少?
(2)若取A点电势为零,小球在B点的电势能、电势分别为多大?
(3)小球到B点时速度为多大?绳子张力为多大?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000746610285.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250007466261111.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250007466412884.jpg)
(1)小球到达最低点B的过程中重力势能、电势能分别变化了多少?
(2)若取A点电势为零,小球在B点的电势能、电势分别为多大?
(3)小球到B点时速度为多大?绳子张力为多大?
(1)-4.5×10-3 J;3×10-3 J(2)3×10-3 J;1.5×103 V(3)1 m/s;5×10-2 N
试题分析:(1)重力势能变化量:ΔEp=-mgl=-4.5×10-3 J
电势能变化量:ΔEp电=Eql=3×10-3 J
(2)小球在B点的电势能:Ep=3×10-3 J
小球在B点的电势:Ep=φBq 解得:φB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000746672720.png)
(3)A→B由动能定理得:mgl-Eql=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000746688343.png)
在B点对小球:FT-mg=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000746704616.png)
解得: FT=5×10-2 N
![](http://thumb2018.1010pic.com/images/loading.gif)
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