题目内容


如图8所示,Q为固定的正点电荷,AB两点在Q的正上方与Q相距分别为h和0.25h,将另一点电荷从A点由静止释放,运动到B点时速度正好又变为零.若此电荷在A点处的加速度大小为g,此电荷在B点处的加速度大小为________;方向________;AB两点间的电势差(用Qh表示)为________.


答案 3g 方向竖直向上 -

解析 这一电荷必为正电荷,设其电荷量为q,由牛顿第二定律,在A点时mgm·g.

B点时mgm·aB

解得aB=3g,方向竖直向上,q.

AB过程,由动能定理mg(h-0.25h)+qUAB=0,

UAB=-.

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