A parallel plate capacitor is charged and then isolated. Now a dielectric slab is Introduced in it. Which of the following quantities will remain constant ?

A parallel plate capacitor is charged and then isolated. Now a dielect

1.	A parallel plate capacitor is charged and then isolated. Now a dielectric slab is Introduced in it. Which of the following quantities will remain constant ?
Answer» Electric charge Q Potential difference V Capacitance C Energy U Solution :When dielectric SLAB is introduced in a parallel plate capacitor, its capacitance C = CK, where K = dielectric constant of a slab. But when C. is INCREASED accroding to `V = (Q)/(C) ` the potential V is DECREASED . Hence energy `U = (1)/(2)C^(1)V^(2)` also changes so charge Q remains constant.

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A parallel plate capacitor is charged and then isolated. Now a dielectric slab is Introduced in it. Which of the following quantities will remain constant ?

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