Use the worked example above to help you solve this problem. A parallel-plate ca

Question

Use the worked example above to help you solve this problem. A parallel-plate capacitor has an area A = 2.30 times 10^-4 m^2 and a plate separation d = 1.20 times 10^-3 m. (a) Find its capacitance. F (b) How much charge is on the positive plate if the capacitor is connected to a 3.00 V battery? C (e) Calculate the charge density on the positive plate, assuming the density is uniform. C/m^2 (d) Calculate the magnitude of the electric field between the plates. V/m Two plates, each of area 2.90 times 10^-4 m^2, are used to construct a parallel-plate capacitor with capacitance 1.40 pF. (a) Find the necessary separation distance. d = m (b) If the positive plate is to hold a charge of 5.30 times 10^-12 C, find the charge density. sigma = C/m^2 (c) Find the electric field between the plates. E = N/C (d) What voltage battery should be attached to the plate to obtain the preceding results? V = V

Explanation / Answer

PART 1)

a) capacitance

C = AEo/d = 2.3*10^-4*8.854*10^-12 / (1.2*10^-3) = 1.697*10^-12 Farad

b) charge on posotive plate

Q = C*v = 1.697*10^-12*3 = 5.091*10^-12 C

c) charge density on plate

d = Q/A = 5.091*10^-12/(2.3*10^-4) = 2.21*10^-8 C/m^2

d) electric field betweem plates

E = v/d = 3.0 / (1.2*10^-3) = 2.5*10^3 V/m

PART 2)

a) C = AEo/d

=> d = AEo/C = 2.9*10^-4*8.854*10^-12/(1.4*10^-12) = 1.834*10^-3 m

b) charge density = Q/A = 5.3*10^-12/(2.9*10^-4) = 1.827*10^-8 C/m^2

c) E = v*d = Q*d/C = 5.3*10^-12*1.834*10^-3 / (1.4*10^-12) = 6.94*10^-3 N/C

d) Q = C*v

=> v = Q/C =    5.3*10^-12 / (1.4*10^-12) = 3.78 volt

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