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? 若场强大小不变,方向变为水平向左,平衡时悬线的拉力T? 答案:解:(1)物体受力分析如图:由平衡条件: T=G+F 3mg=mg+Eq 解得:E= 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; 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.Find out the mistakes and correct them. 找出错误并改正。( ) 1. I have to put on my boots on a rain day. A B C ( ) 2. Li Ming is writing an letter in the corner. A B C ( ) 3. The peas are in the stove. A B C ( ) 4. I'm a 12 year old boy. A B C( ) 5. Danny, you should sing soft. A B C