help please :) You are asked to design a parallel-plate capacitor having a capac

Question

help please :)

You are asked to design a parallel-plate capacitor having a capacitance of 1.04 F and a plate separation of 1.55 mm. Calculate the required surface area of each plate. The permittivity of free space is 8.85 times 10-12 C2/N m2 Answer in units of m2 Each plate on a 3876 pF capacitor carries a charge with a magnitude of 1.62 times 10-8 C. What is the potential difference across the plates when the capacitor has been fully charged? Answer in units of V If the plates are 6.63 times 10-4 m apart, what is the magnitude of the electric field between the two plates? Answer in units of V/m Consider the capacitor network. Find the equivalent capacitance between points a and b. Answer in units of mu F

Explanation / Answer

C= 1.04F

d= 1.55mm

C= eps*A/d

A = 1.04*1.55*0.001/8.85*10^-12 = 1.82*10^8 m^2

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C = 3876 pF

Q = 1.62*10^-8C

Q=CV

V = Q/C = 4.18 V

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d=6.63*10^-4 m

E=V/d = 6304.7 V/m

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C1 = 8.6*4.2/(8.6+4.2) = 2.82

Ceq = 2.82+6.1 +6 = 14.92 muF

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