题目内容
如图1、2、3、4虚线是正点电荷形成电场中的等势线,相邻等势线的电势差大小均为10V,且令
。某一带电量q=0.1C的带电粒子,沿图中实线运动轨迹,先后通过A、B、C三点,经过A点时,带电粒子的动能EkA=10 J,则关于动能、电势能、电势,以及电场力做功,下列说法正确的是( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241608226245533.jpg)
A、EkB=2 J,WAB=-8 J,![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160822640639.png)
B、EkC=9 J,WBC=1 J,
C、EkC=1 J,WAC=1 J,![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160822749605.png)
D、从A到B电场力做正功,从B到C电场力做负功
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160822484457.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241608226245533.jpg)
A、EkB=2 J,WAB=-8 J,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160822640639.png)
B、EkC=9 J,WBC=1 J,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160822749605.png)
C、EkC=1 J,WAC=1 J,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160822749605.png)
D、从A到B电场力做正功,从B到C电场力做负功
B
试题分析:相邻等势线的电势差大小均为10V,而沿电场线方向电势减小,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160823139590.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160823404617.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160823623825.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160823841591.png)
从B运动到C,电场力做正功,电势能减小,动能增大,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160824044771.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160824293906.png)
从A运动到C过程中,电场力做负功,故
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160824590605.png)
故选B,
点评:做本题需要注意电场力做功情况,先根据等势面判断电场线的方向,根据正电荷电场规律判断电场力做功情况
![](http://thumb2018.1010pic.com/images/loading.gif)
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