A 2.0-nF parallel plate capacitor with a sheet of mylar (k = 3.1)filling the space between the plates is charged to a potential difference of 90 V and is then disconnected.How much work would be required to completely remove the sheet of mylar from the space between the two plates?

Question

Quizplus · Accepted Answer

The work required to remove the dielectric is equal to the change in energy of the capacitor. Initially, the energy is (1/2)CkV^2, and after removing the dielectric, it is (1/2)CV^2. The difference is (1/2)C(k-1)V^2 = (1/2)(2.0 * 10^-9 F)(3.1 - 1)(90^2) = 1.7 * 10^-5 J.

A 20-NF Parallel Plate Capacitor with a Sheet of Mylar (K