Question

如圖所示，在光滑絕緣豎直細桿上，套有一小孔的帶電小球，小球質(zhì)量為m，帶電荷量為﹣q，桿與以正電荷Q為圓心的某一圓周交于BC兩點，小球從A點無初速釋放，AB=BC=h，小球滑到B點時速度大小為，已知q遠小于Q，求：                                                                                                                                

①AB兩點間的電勢差；                                                                                              

②AC兩點間的電勢差；                                                                                              

③小球滑到C點時速度的大�。�                                                                                  

                                                                                                             

                                                                                                                                   

Answer 1

解：（1）小球由A到B重力和電場力做功，由動能定理得         

mgh+W_AB=mv²﹣0                          

代入數(shù)據(jù)解得W_AB=mgh 

故AB兩點間的電勢差為：

（2）由電勢差的定義得：

U_AC=U_AB=

代入數(shù)據(jù)解得：

U_AC=

（3）小球由A到C由動能定理得                

2mgh+W_AB=mv_c²﹣0                          

代入數(shù)據(jù)解得：

v_C=

答：（1）AB兩點間的電勢差為；

（2）A、C兩點間的電勢差為；

（3）小球下滑到C點時速度大小為．