Dielectrics: A parallel-plate capacitor has a capacitance of 10 mF and is charged with a 20-V power supply. The power supply is then removed and a dielectric material of dielectric constant 4.0 is used to fill the space between the plates. What is the voltage now across the capacitor?

Question

Quizplus · Accepted Answer

The capacitance of the capacitor with the dielectric material between the plates is given by:
C' = κC
where κ is the dielectric constant.
Thus, the new capacitance is:
C' = 4.0 × 10 mF = 40 mF
The charge on the capacitor remains the same after the dielectric is inserted since charge is conserved. Therefore, the voltage across the capacitor with the dielectric is given by:
V' = Q/C' = Q/(κC)
Since Q/C = 20 V,
V' = 20 V/4.0 = 5.0 V.

Dielectrics: a Parallel-Plate Capacitor Has a Capacitance of 10 MF