Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1

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Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1

.80 millimeters thick, with a dielectric constant of K=3.60. The resultant electric field in the dielectric is 1.20×106 volts per meter.
(A) Compute the magnitude of the charge per unit area sigma on the conducting plate.
(B) Compute the magnitude of the charge per unit area sigma1 on the surfaces of the dielectric.
(C) Find the total electric-field energy U stored in the capacitor.

Physics

1 answer:

ValentinkaMS [3K] · Answer 1 · 2026-03-28T05:41:09+03:00

A). σ = 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$ B). $\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$ C). $U=10.322$ J. To find the charge per unit area for a conducting plate, we have that E equals 1.2 x $10^{6}$ V/m, and the permittivity of free space is 8.85 x $10^{-12}$ $C^{2}$ /N- $m^{2}$ with a dielectric constant k of 3.6. Calculating this gives σ = 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$ . For the magnitude of charge on the dielectric surface, it is given by C/ $m^{2}$ . Lastly, with an area A = 2.5 $cm^{2}$ , the total energy stored in the capacitor is J.