Question
Asked Nov 6, 2019
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A parallel-plate air-filled capacitor having area 49.0 cm2 and plate spacing 0.800 mm is charged to a potential difference of 820 V. Find (a) the capacitance, (b) the magnitude of the charge on each plate, (c) the stored energy, (d) the electric field between the plates, (e) the energy density between the plates.

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Expert Answer

Step 1

Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts  you’d like answered.

Given,

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Area of capacitor plate, A = 49 cm2 = 49x10 m2 Plate spacing,d=0.80 mm =0.80x 103 m Potential Difference,V 820 V

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Step 2

Part (a):

Capacitance of capacitor can be calculated as,

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C= c(N'm2) where, Permittivity of dielectric 8.85 x 1012 C Capacitance of capacitor

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Step 3

By substituting the ...

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C= (8.85x10-")(49x10*) (0.80x10) C= C 5.42x10 F C 54.2x1012F (1 pico-10 C 54.2p F

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Science

Physics

Electrostatic Potential and Capacitance

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