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3 years ago

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A capaitor has two parallel plates sepearted by 2mm and is connected across a 50V battery. i. What is the electric field between

the plates? ii. What is the surface charge density? iii. How much charge is stored on each plate if the area is 0.1m. iv. Calculate the capacitance. v. How much energy is stored in this capacitor? vi. What is the energy density of the electric field in the capacitor?

1 answer:

Oduvanchick [21]3 years ago

8 0

Answer:

i E=V/d=50/2*10^-3=25*10^3 N/C

ii It is a (+) and (-)

iii C=εA/d

C=12.56*10^-8 * 0.1/2*10^-4

C=62.83 μF

Q=CV=50*6.283*10^-6

Q=314 μC

iv E=0.5 QV

=0.5(50*314*10^-6)

=7850 μJoule

Explanation:

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