题目内容
如图所示,一根绝缘杆长为L,两端分别带等量异种电荷,电荷量绝对值为Q,处在场强为E的匀强电场中,杆与电场线的夹角为60°,若是杆沿顺时针方向转过60°(以杆上某一点为圆心转动),则下列叙述正确的是( )![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029194322009146146/SYS201310291943220091461006_ST/images0.png)
A.电场力不做功,两电荷的电势能不变
B.电场力做的总功为
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029194322009146146/SYS201310291943220091461006_ST/0.png)
C.电场力做的总功为-
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029194322009146146/SYS201310291943220091461006_ST/1.png)
D.电场力做的总功大小跟转动轴有关
【答案】分析:杆沿顺时针方向转过60°,电场力对两电荷都做正功,电势能都减少.根据W=qEd计算电场力做功,其中d是沿电场方向两点间的距离.
解答:解:A、+Q所受电场力水平向右,-Q所受电场力水平向左,当杆沿顺时针方向转过60°时,电场力对两个电荷都做正功,两电荷的电势能都减小.故A错误.
B、C,电场力对正电荷所受的功W1=qE
(1-cos60°)=
,电场力对正电荷所受的功W2=qE
(1-cos60°)=
,电场力做的总功为W=W1+W2=
.由于电场力做正功,两个电荷的电势能减少.故B正确,C错误.
D、由上得到总功W=
,可见,总功与跟转动轴无关.故D错误.
故选B
点评:本题是电偶极子,电场力对两个电荷做的总功大小跟转动轴无关.基础题.
解答:解:A、+Q所受电场力水平向右,-Q所受电场力水平向左,当杆沿顺时针方向转过60°时,电场力对两个电荷都做正功,两电荷的电势能都减小.故A错误.
B、C,电场力对正电荷所受的功W1=qE
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029194322009146146/SYS201310291943220091461006_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029194322009146146/SYS201310291943220091461006_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029194322009146146/SYS201310291943220091461006_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029194322009146146/SYS201310291943220091461006_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029194322009146146/SYS201310291943220091461006_DA/4.png)
D、由上得到总功W=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029194322009146146/SYS201310291943220091461006_DA/5.png)
故选B
点评:本题是电偶极子,电场力对两个电荷做的总功大小跟转动轴无关.基础题.
![](http://thumb2018.1010pic.com/images/loading.gif)
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