A 2.0-nF parallel plate capacitor with a sheet of mylar (k = 3.1)filling the space between the plates is connected to a 90 V battery.While it is still connected to the battery,the sheet of mylar is completely removed from the space between the two plates.Determine the change in the charge on one of the plates.

Question

Quizplus · Accepted Answer

The capacitance of a parallel plate capacitor filled with a dielectric material is given by $C = kC_0$, where $C_0$ is the capacitance without the dielectric, and $k$ is the dielectric constant. Initially, the capacitance with mylar is $C = 3.1 \times 2.0\,nF = 6.2\,nF$. The charge on the capacitor is $Q = CV = 6.2\,nF \times 90\,V = 558\,nC$. After removing the mylar, the capacitance returns to $2.0\,nF$, but since the capacitor is still connected to the battery, the voltage remains the same, so the new charge is $Q' = 2.0\,nF \times 90\,V = 180\,nC$. The change in charge is $\Delta Q = Q' - Q = 180\,nC - 558\,nC = -378\,nC = -0.378\,\mu C$, which is closest to choice B, -1.2 x $10^{-7}C$.

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