A、B两带电的小球,A固定不动,B的质量为m.在库仑力的作用下,B由

A、B两带电的小球,A固定不动,B的质量为m.在库仑力的作用下,B由静止开始运动.已知初始时,A、B间的距离为d,B的加速度为a,经过一段时间后,B的加速度变为,此时,A、B间的距离应为多少?

答案

解析:A、B两球的加速度都是由静电力作用产生的,万有引力忽略不计.设A、B两球带电荷量分别为qa、qb,则静电力为F=k

由牛顿第二定律可知,开始时的加速度为a=,当加速度改变后,=.联立解得:x=2d.

答案:x=2d

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