（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。
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。
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