问题 计算题

(13分)如图所示,一质量为m的带电量为q的小球,用长为L的绝缘细线悬挂在水平向右的匀强电场中,静止时悬线与竖直方向成θ=370角。(重力加速度为g,cos370=0.8,sin370=0.6)

(1)判断小球带何种电荷。

(2)求电场强度E。

(3)在图示位置,若将细线突然剪断,小球做何种性质的运动?求加速度a的大小。

答案

题目分析:⑴由图可知,小球受到的电场力方向向左,电场方向向右,所以小球带负电。

⑵小球受三个力作用处于平衡状态,受力分析图见上图,有tan370="Eq/mg"

可得:E=3mg/4q         

⑶从图示位置将线剪断后,小球只受重力和电场力的作用,所以小球将做匀加速直线运动

小球受的合力为F=mg/cos370 根据牛顿第二定律得:F="ma"

解得:a=1.25g  

点评:学生要画出小球的受力示意图,是解本题的关键。

完形填空
完形填空。
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