如图所示,在竖直平面内,光滑绝缘直杆AC与半径为R的圆周交于B、C

如图所示,在竖直平面内,光滑绝缘直杆AC与半径为R的圆周交于B、C两点,在圆心处有一固定的正点电荷,B为AC的中点,C点位于圆周的最低点。现有一质量为m、电荷量为-q套在杆上的带负电小球从A点由静止开始沿杆下滑。已知重力加速度为g,A点距过C点的水平面的竖直高度为3R,小球滑到B点时的速度大小为。求:

(1)小球滑至C点时的速度的大小;

(2)A、B两点的电势差

答案

(1) vc=(2)=/q=-mgR/2q


解析:

(1) B—C  3mgR/2=mvc2/2-mvB2/2

           vc=

A—B:  3mgR/2+=mVB2/2-0    =mgR/2    =/q=-mgR/2q

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