题目内容
一质量为4.0×10-15 kg、电荷量为2.0×10-9 C的带正电质点,以4.0×104 m/s的速度垂直于电场方向从a点进入匀强电场区域,并从b点离开电场区域,离开电场时的速度为5.0×104 m/s.由此可见,电场中a、b两点间的电势差φa-φb=_______V.带电质点离开电场时,速度在电场方向的分量为_________m/s.(不考虑重力作用)
9.0×102 3.0×104
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241200470301938.jpg)
图13-9-12
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241200470301938.jpg)
图13-9-12
qUab=
mv2-
mv
,代入数据得Uab=9.0×102 V
v⊥=
=3.0×104 m/s.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120047108226.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120047108226.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120047202193.gif)
v⊥=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120047217317.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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