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# parallel plates capacitor and its capacitance

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Two parallel plates are .005m apart and are each 2m2 in area. The plates are in vacuum and an electric potential difference of 10,000V is applied across them.

1) Find the:
a)capacitance,
b)the charge on each plate
c)the electric field intensity in the space between, and
d) the stored energy.

2) If a dielectric material with a dielectric constant &#61547; = 80.4 is inserted into the gap between the plates, with the electric potential running the same, find the
a) new capacitance
b) charge on each plate
c) electric field intensity, and
d) energy stored

##### Solution Summary

The solution is comprised of detailed explanations of parameters related to parallel plates capacitor, such as capacitance, charges on the plates, voltage, electric field in between the plates, energy stored. It also explains how these parameters change when a dielectric material is inserted between the plates.

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Two parallel plates are .005m apart and are each 2m2 in area. The plates are in vacuum and an electric potential difference of 10,000V is applied across ...

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