Question

The electric field between the plates of a paper-separated (K=3.75) capacitor is 8.23×104 V/m . The plates are 1.75 mm apart, and the charge on each plate is 0.705 μC . Determine the capacitance of this capacitor. Determine the area of each plate

Answer 1

Part A.

We know that charge across capacitor is given by:

Q = C*V

C = Q/V

Q = Charge = 0.705*10^-6 C

V = Potential across plates = E*d

V = (8.23*10^4 V/m)*(1.75*10^-3 m)

V = 144.0 V

C = Capacitance of capacitor = 0.705*10^-6/144.0

C = 4.90*10^-9 F

C = 4.90 nF

Part B.

Now capacitance is also given by:

C = k*e0*A/d

A = C*d/(e0*k)

e0 = permittivity of free space = 8.854*10^-12

k = dielectric constant of paper = 3.75

So,

A = 4.90*10^-9*1.75*10^-3/(8.854*10^-12*3.75)

A = 0.258 m^2 = Area of each plate

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