Question

A parallel-plate capacitor with air between its plates is connected to an 80.0 V battery. When fully charged, the capacitor has an energy of 130 nJ. Without disconnecting the battery, a slab of dielectric is inserted between the plates of the capacitor, fully filling the gap. The energy stored in the capacitor is now 410 nJ.

(c) What is the dielectric constant of the dielectric?

(d) What is the charge on the dielectric-filled capacitor?

Answer 1

Since, , where k is the dielectric constant.

And using Energy of parallel plate capacitor ,

We have,

Energy of a parallel plate capacitor can also be written as by replacing C in using equation Q = CV of the parallel plate capacitor.

Therefore, Energy of the dielectric filled parallel plate capacitor

where Q is the charge on the dielectricfilled parallel plate capacitor.

Hence, charge on the parallel plate capacitor filled with dielectric is ,