Acharge of 2.0 × 10–10 c is to be stored on each plate of a parallel-plate capacitor having an area of 650 mm2 (1.0 in.2 ) and a plate separation of 4.0 mm (0.16 (a) what voltage is required if a material having a dielectric constant of 3.5 is positioned within the plates? (b) what voltage would be required if a vacuum were used? (c) what are the capacitances for parts (a) and (b)? (d) compute the dielectric displacement for part (a). (e) compute the polarization for part (a).

a) 43.7 V b) 153 V c) a) 5 pf b) 1.43 pf d) 0.34 μC/m²

              e) 0.24 μC/m²

Explanation:

a) By definition, the capacitance is given by the following relationship between the charge on one of the plates, and the voltage between them:

C= Q/V

For a parallel-plate capacitor, C is also obtained as a constant given by the geometry and the material of the dielectric, as follows:

C= εA/d, where A is the area of one of the plates, and d the separation between them.

ε refers to the permitivitty of the dielectric material, and is related with the dielectric constant of the material, as follows:

ε = ε₀ .εr , where ε₀ = 8.85. 10⁻¹² farad/m, and εr = 3.5

Equating both expressions for C, and solving for V, we have:

V = Q. d / ε₀ εr A

Replacing for the values, we finally arrive to this value for V:

V = 43.7 V

b) If we look to the expression for C, that depends on ε, for vacuum, εr=1, so same considerations apply, and we find the new value for V, as follows:

V = 43.7 x 3.5 = 153 V

c) Applying the definition of capacitance, and assuming that charge on the plates remains constant, as voltage can be obtained from a) and b) above. we get:

C1 = 5 pF

C2 = 1.43 pF

d) Applying Gauss' Law, and the definition of electric flux, it can be showed that there exist a relationship between the dielectric displacement D and the electric field E, as follows:

D = εE =ε₀εrE

In a capacitor, the electric field (assumed constant) is related with the voltage between plates, as follows:

E = V/d where d is the separation between plates.

So we can find D, replacing E by V/d, so we have:

D = (ε₀εrV)/d=8.85. 10⁻¹² F/m.3.5. 43.7 V / 4.10⁻³ m = 0.34 μC/m²

e) When a dielectric is put in a place where an electric field is present, it is produced a relative reorientation of his atoms, causing an internal electric field that opposes to the external one, reducing his effect overall.

One way to explain this, is due to the existence of an internal field, representented by a vector called polarization, that is related with the dielectric displacement and the electric field as follows:

D = ε₀E + P

P = D - ε₀E

With the values for D and E from d) above, we get P:

P = 0.24 μC / m²