A parallel plate capacitor is created by placing two large square conducting pla

Question

A parallel plate capacitor is created by placing two large square conducting plates of length and width 0.50 meters facing each other, separated by a 1.00-centimeter gap. A battery is connected to the two plates so that a charge density of =500C/m2 is stored. The battery is then removed from the plates. The plates are close together so electric field is uniform and constant. The electric potential difference between the plates is approximately

225,000 V

327,000 V

565,000 V

814,000 V

225,000 V

327,000 V

565,000 V

814,000 V

Explanation / Answer

for electric field using gauss' law

integral(E.ds) = Qin/e0

E*A = sigma*A/e0

E = sigma/e0

potential difference = V =E*d

V = sigma*d/e0

d = 1.00 cm = 0.01 m

V = 500*10^-6*0.01/(8.85*10^-12) = 564971.5

Potential difference = 565000 V

Option C is correct.

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