5t s (aue Thurs 2tat11:59PM) Parallel Timer Notes Evaluate Feedback plate capaci

Question

5t s (aue Thurs 2tat11:59PM) Parallel Timer Notes Evaluate Feedback plate capacitor in air and in water Two parallel plates, each of area 1.82 cm2, are separated by 5.70 mm. The space between the plates is filled with air. A voltage is applied between the plates. Calculate the magnitude of the electric field between the plates of 3.05 S.omt Ansaer Tries 0/20 Calculate the amount of the electric charge stored on each plate. Submit Answ Tries 0/20 Now distilled water is placed between the plates and magnitude of charge stored on each plate in this case. (Use K 80.0 for the dielectric constant of water Sutmit Ansae Tries 0/20 ( Post Discussion Send Feedback

Explanation / Answer

a) E = V / d = 3.05 / (5.70 * 10-3)

magnitude of electric field = 535.1 V/m

b) C = e0 A / d

= (8.85 * 10-12 * 1.82 * 10-4) / (5.70 * 10-3) = 2.826 * 10-13 F

Q = C V = 2.826 * 10-13 * 3.05

electric charge stored on each plate = 8.62 * 10-13 C

c) k = 80

Q' = k Q = 80 * 8.62 * 10-13

magnitude of charge = 6.89 * 10-11 C

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