题目
把质量0.2g的带负电的小球A用绝缘线挂起,现将带电为Q=4.0×C的带电小球B靠近A,两球在同一水平面上,相距30cm,此时A球悬线与竖直方向成45°而平衡,如图所示,求: ①B球受到的库仑力大小; ②A球电量; ③B在球A平衡处的场强. 答案:(1)2×N (2)5×10-7C (3)4×103N/C 解析: (1)T=mg/cos f=Tsin ∴f=mg=2×(N) My uncle often teach me _________ kind to others.A.beB.to amC.to beD.become