Worksheet: Dielectric Constant
In this worksheet, we will practice calculating the capacitance of a capacitor given the dielectric constant of the material separating the capacitor plates.
A parallel-plate capacitor has square plates that are 8.00 cm on each side and 3.80 mm apart. The space between the plates is completely filled with two square slabs of dielectric, each 8.00 cm on a side and 1.90 mm thick. One slab is Pyrex glass with a dielectric constant and the other slab is polystyrene with a dielectric constant . If the potential difference between the plates is 86.0 V, find how much electrical energy can be stored in this capacitor.
A parallel-plate capacitor with only air between its plates is charged by connecting the capacitor to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates.
A voltmeter reads 45.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of the material?
What will the voltmeter read if the dielectric is now pulled out from between the capacitor plates so that dielectric material fills only one-third of the space between the plates?
The dielectric constant of Teflon is 2.1. A Teflon-filled, parallel-plate capacitor has a plate area of 50.0 cm2. The spacing between these plates is 0.50 mm, and the capacitor is connected to a 200 V battery.
Find the free charge on the capacitor plates.
Find the electrical field in the dielectric.
Find the induced charge on the dielectric’s surfaces.
A parallel plate capacitor has plates of area 1.5 m2. The plates are separated by 0.020 mm of neoprene rubber that has a dielectric constant of 6.7.
What is the capacitance of the capacitor?
- A 4.4 µF
- B 6.0 µF
- C 5.3 µF
- D 3.0 µF
- E 3.7 µF
What charge is stored in the capacitor when a 9.0-V-potential difference is applied across it?
- A C
- B C
- C C
- D C
- E C