如图所示，A为带+Q的金属板，沿金属板的垂直平分线距板r处有一用绝缘线悬挂于O点的质量为m、电量大小为q 的小球由于受水平向右的电场力使悬线偏转θ角而静止。则带电金属板在小球所在处产生的场强方向为 ，场强大小为 。
mgtanθ/q
题目分析：小球受到重力、绳子的拉力和电场力处于平衡状态，分析小球的受力可知电场力为mgtanθ，由F=qE可知E= mgtanθ/q，电场力方向向右，正电荷所受电场力方向与场强方向相同，由此可知场强方向水平向右
点评：本题为典型的受力平衡问题，对研究对象进行受力分析可求解，难度较小
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| Here are my favorite photos. 1 the first one, you can find it's 2 . I'm wearing a T-shirt and sitting on the grass. The 3 one is a photo of my family. I'm 4 computer games. My grandfather is playing Erhu. My father is 5 a newspaper. My grandmother is talking about the TV show 6 my sister. And my mother is 7 the table. I love this photo very much. In the third one, there are two 8 . They are my friends, Tim and his brother. They are playing basketball. In the last one, I'm singing and my friends are dancing. 9 I see these photos, I'm very happy. They make me think of the happy 10 . I will keep the photos all my life. | |||
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