A parallel plate capacitor of capacitance C0 has plates of area A with separation b between them. When the capacitor is connected to a battery of voltage ÄV, it has a charge of magnitude Q on its plates. It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant ê = 3. After the dielectric material is added, what is the potential difference between the plates?

Question

Quizplus · Accepted Answer

When the dielectric is added, the capacitance increases by a factor of the dielectric constant (ê = 3), but the charge remains the same since the capacitor is disconnected. The potential difference is reduced by the same factor, so it becomes ÄV/3.

A Parallel Plate Capacitor of Capacitance C0 Has Plates of Area