Question #166126

A parallel plate capacitor has an area of 5.00 cm 2 and a capacitance of 3.50 pF. The

capacitor is connected to a 12.0 V battery. After the capacitor is completely charged up, the

battery is removed

a. What must be the separation of the plates and how much energy is stored in the electric

field between the plates?

Expert's answer

The capacitance of a parallel-plate capacitor is defined as

where

- A is the area of the plates (5 cm
^{2}=0.0005 m^{2}) - d is the plate separation (m)
- "\\epsilon_0" is the vacuum permittivity (~8.85*10
^{-12}F/m) - "\\epsilon_r" is the relative permittivity of air (1.0006)

Then the separation of the plates:

The amount of energy stored between the plates of a parallel-plate capacitor:

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