Question

质量为m，电量为+q的带电小球用绝缘细线悬挂在O点，现加一竖直向下的匀强电场，悬线所受拉力为3mg．                                                                                                                                   

（1）求所加电场的场强E为多大？                                                                               

（2）若场强大小不变，方向变为水平向左，平衡时悬线的拉力T？                                    

                                                                                                                              

                                                                                                                                     

Answer 1

（1）物体受力分析如图：由平衡条件：

T=G+F

3mg=mg+Eq

解得：E=

（2）平衡时受力分析如图：T=，

设T的方向与水平方向的夹角为θ，则，解得：，

所以方向与水平方向成向右上．

答：（1）求所加电场的场强E为；

（2）若场强大小不变，方向变为水平向左，平衡时悬线的拉力大小为，方向与水平方向成向右上．