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In this lesson, we will learn how to calculate the capacitance of two parallel charged plates given their surface area and the distance between them.
The plates of an empty parallel-plate capacitor of
capacitance 5.0 pF are 2.0 mm apart. What is the area of
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?
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
What is the capacitance of the shelves?
What is the potential difference across the shelves?
How much energy is stored in the electric field between the shelves?
An 8.0 pF vacuum capacitor has a plate area of 0.070 m2. What is the
separation between its plates?
Two parallel conducting plates, each of cross-sectional area 580 cm2, are 3.2 cm apart and
electrons are transferred from one plate to the other.
What is the charge density on each plate?
What is the magnitude of the electric field between the plates?
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
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
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
An empty parallel plate capacitor has a capacitance of 80 µF. How much charge must
leak off its plates before the voltage across them is reduced by 250 V?
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?
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