sat physics (electric potential) help!a parallel plate capacitor is charged to a potential difference of V,this results in a charge of +Q on one plate and a charge of -Q on the other.The capacitor is disconnected from the charging source,and a dielec

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sat physics (electric potential) help!a parallel plate capacitor is charged to a potential difference of V,this results in a charge of +Q on one plate and a charge of -Q on the other.The capacitor is disconnected from the charging source,and a dielec
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sat physics (electric potential) help!a parallel plate capacitor is charged to a potential difference of V,this results in a charge of +Q on one plate and a charge of -Q on the other.The capacitor is disconnected from the charging source,and a dielec
sat physics (electric potential) help!
a parallel plate capacitor is charged to a potential difference of V,this results in a charge of +Q on one plate and a charge of -Q on the other.The capacitor is disconnected from the charging source,and a dielectric is then inserted.what happens to the potential difference and the stored electrical potential energy?
用decrease increase,unchange来回答
potential difference 和 stored elextrical potential energy是什么区别吗?
which points in the uniform electric field between the plates of the capacitor lie on the same equipotential?(图是一个capacitor,上面有4个点,两个是和capacitor平行,还有两个是和capacitor垂直) 答案是那两个和capacitor平行的点,为什么?他不是说是uniform electri field吗?那不应该是4个点的电压都该是一样的?
由于本人理解能力有限,就当我是一个初中生= =

sat physics (electric potential) help!a parallel plate capacitor is charged to a potential difference of V,this results in a charge of +Q on one plate and a charge of -Q on the other.The capacitor is disconnected from the charging source,and a dielec
decrease decrease
当电介质（dielectric）插入电容时,会增大电容,而且电容已经断离电路（也就是说电荷恒定,因为不会再有电荷转移）,根据公式C=Q/U C增大,Q不变,U（电势差）变小
stored electrical potential energy = Q^2/(2*C) C增大 Q不变 所以energy变小

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