1.

A parallel plate capacitor, with a slab of dielectric constant k between its plates, has a capacitance C. It is charged to a potential V. Now the dielectric slab is first brought out of the capacitor and is then introduced again. The work done in this process will be

Answer»

`(K-1)(CV^2)/(2)`
`(CV^2(k-1))/(k)`
`(k-1)CV^2`
0

Answer :D


Discussion

No Comment Found

Related InterviewSolutions