Question

如图所示，在光滑绝缘竖直细杆上，套有一小孔的带电小球，小球质量为m，带电荷量为﹣q，杆与以正电荷Q为圆心的某一圆周交于BC两点，小球从A点无初速释放，AB=BC=h，小球滑到B点时速度大小为，已知q远小于Q，求：                                                                                                                                

①AB两点间的电势差；                                                                                              

②AC两点间的电势差；                                                                                              

③小球滑到C点时速度的大小．                                                                                  

                                                                                                             

                                                                                                                                   

Answer 1

解：（1）小球由A到B重力和电场力做功，由动能定理得         

mgh+W_AB=mv²﹣0                          

代入数据解得W_AB=mgh 

故AB两点间的电势差为：

（2）由电势差的定义得：

U_AC=U_AB=

代入数据解得：

U_AC=

（3）小球由A到C由动能定理得                

2mgh+W_AB=mv_c²﹣0                          

代入数据解得：

v_C=

答：（1）AB两点间的电势差为；

（2）A、C两点间的电势差为；

（3）小球下滑到C点时速度大小为．