Worksheet: Capacitance of a Parallel Plate Capacitor
In this worksheet, we will practice calculating the capacitance of two parallel charged plates given their surface area and the distance between them.
An air-filled (empty) parallel-plate capacitor is made from two square plates that are 25 cm on each side and 1.0 mm apart. The capacitor is connected to a 50-V battery and fully charged. It is then disconnected from the battery and its plates are pulled apart to a separation of 2.00 mm.
What is the capacitance of this new capacitor?
What is the charge on each plate?
What is the electrical field between the plates?
An anxious physicist worries that the two metal shelves of a wood frame bookcase might obtain a high voltage if charged by static electricity, perhaps produced by friction. The shelves have an area of 1.00 m2 and are separated by a distance of 0.200 m. A total charge of 2.00 nC is placed on the two shelves.
What is the capacitance of the shelves?
- A F
- B F
- C F
- D F
- E F
What is the potential difference across the shelves?
How much energy is stored in the electric field between the shelves?
- A J
- B J
- C J
- D J
- E J
Two parallel conducting plates, each of cross-sectional area 580 cm2, are 3.2 cm apart and uncharged. electrons are transferred from one plate to the other.
What is the charge density on each plate?
- A C/m2
- B C/m2
- C C/m2
- D C/m2
- E C/m2
What is the magnitude of the electric field between the plates?
- A N/C
- B N/C
- C N/C
- D N/C
- E N/C
A capacitor is made from two flat parallel plates placed 1.9 mm apart. When a charge of 0.040 µC is placed on the plates, the potential difference between them is 360 V.
What is the capacitance of the plates?
What is the area of each plate?
What is the charge on the plates when the potential difference between them is 850 V?
What is the maximum potential difference that can be applied between the plates so that the magnitude of electric fields between the plates does not exceed 4.5 MV/m?
A parallel plate capacitor with a capacitance of 8.0 µF is charged with a 0.5 V battery, after which the battery is disconnected. What is the minimum work required to increase the separation between the plates by a factor of 5?