A parallel-plate capacitor with area 0.700 m^2 and plate separation of 4.00 mm i

Question

A parallel-plate capacitor with area 0.700 m^2 and plate separation of 4.00 mm is connected to a 18.00-V battery. What is the capacitance? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. F How mu Ch charge is stored on the plates?. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. C What is the electric field between the plates? Find the magnitude of the charge density on each plate. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. C/m^2

Explanation / Answer

(a)

capacitance = epslon A/d = 8.84 * 10^-12 * 0.7 / 4*10^-3 = 1.547nF

(c)

Q = C*V = 1.547nC * 18 = 27.846 nc

(c)

electric field = V/d = 18/4mm = 4500 V/m

(d)

charge density = Q/A = 27.846nC / 0.7 = 39.78 nC/m^2

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