[题目]平行板电容器的两极板A.B接于电池两极.一带正电小球悬挂在电容器内部.闭合开关S.电容器充电.这时悬线偏离竖直方向的夹角为θ.如图所示.则( )A．保持开关S闭合.带正电的A极板向B极板靠近.则θ角增大B．保持开关S闭合.带正电的A极板向B极板靠近.则θ角不变C．开关S断开.带正电的A极板向B极板靠近.则θ角增大D．开关S断开.� 题目和参考答案——青夏教育精英家教网—

【题目】平行板电容器的两极板A、B接于电池两极，一带正电小球悬挂在电容器内部，闭合开关S，电容器充电，这时悬线偏离竖直方向的夹角为θ，如图所示，则（）

A．保持开关S闭合，带正电的A极板向B极板靠近，则θ角增大

B．保持开关S闭合，带正电的A极板向B极板靠近，则θ角不变

C．开关S断开，带正电的A极板向B极板靠近，则θ角增大

D．开关S断开，带正电的A极板向B极板靠近，则θ角不变

【答案】AD

【解析】球在电场中平衡，则所受电场力、重力及绳的拉力的合力为零。由平衡条件得tanθ＝，故要判断θ的变化，只需判断电场强度E的变化即可。S闭合时，U不变，A向B靠近，d减小，由E＝，可知E增大，θ角增大，故A正确，B错误；S断开，则Q不变，A向B靠近，E不变，则θ角不变，故C错误，D正确。

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