如图所示,水平面中的平行导轨P、Q相距L,它们的右端与电容为C的

如图所示,水平面中的平行导轨P、Q相距L,它们的右端与电容为C的电容器的两极板分别相连,直导线ab放在P、Q上与导轨垂直相交并且沿导轨滑动,磁感应强度为B的匀强磁场竖直向下穿过导轨面。闭合开关S,若发现与导轨P相连的电容器极板上带负电荷,则ab向__        沿导轨滑动(填“左”、“右”);如电容器的带电荷量为q,则ab滑动的速度v=           

 

答案

左                 q/BCL

解析:根据右手定则,可判断出向左运动,根据电容器的带电荷量为q,可求得两端的电压为,根据法拉第电磁感应定律,可得ab滑动的速度v=

 

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