题目内容
如图所示,
是一只理想二极管,水平放置的平行板电容器
内部原有带电微粒
处于静止状态。下列措施下,关于
的运动情况说法不正确的是( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241607549502320.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160753967315.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160754185396.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160754435289.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160754435289.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241607549502320.jpg)
A.保持![]() ![]() ![]() ![]() |
B.保持![]() ![]() ![]() ![]() |
C.断开![]() ![]() ![]() ![]() |
D.断开![]() ![]() ![]() ![]() |
C
试题分析:保持S闭合,电源的路端电压不变.增大A、B板间距离,电容减小,由于二极管的单向导电性,电容器不能放电,其电量不变,由推论
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160757820820.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160757961598.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160758257837.png)
让选错误的,故选C,
点评:本题关键之处要抓住二极管的单向导电性,使得电容器不能放电.对于推论
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824160758257837.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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