Question

A 5.80 muF parallel-plate air capacitor has a plate separation of 5.00 mm and is charged to a potential difference of 400 V. Calculate the energy density in the region between the plates, in J/m^3

Answer 1

Answer:

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Answer 2

Hence the density in the region between the plates is 2.83 x 10^-2 J/m^3

Explanation:

Given data:

• Capacitance of parallel plate = 5.80 muF
• Separation between plates = 5.0 mm
• potential difference = 400 V

Solution:

U = 1/2 εo E^2

U = 1/2 εo [ v / d] ^2

U = 1/2 εo V^2 / d^2

U = 1/2 x 8.85 x 10^-12 x [ 400 / 5 x 10^-3]^2

U = 4.425 x 10^-12 [ 8 x 10^4]^2

U = 4.425 x 64 x 10^-4  J/m^3

U = 283.2 x 10^-4 = 2.83 x 10^-2 J/m^3

Hence the density in the region between the plates is 2.83 x 10^-2 J/m^3