Capacitor with a Dielectric Problems and Solutions2

 Problem #1

In Fig. 01, how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air, and the other is filled with a dielectric for which ε_r = 3.00; both capacitors have a plate area of 5.00 x 10^-3 m² and a plate separation of 2.00 mm.

Fig.1

Answer:

Known:
Potential difference , V = 12.0 V
Dielectric material with relative permitivity ε_r = 3.00
Plate area, A = 5.00 x 10^-3 m²
Plate separation, d = 2.00 mm = 2.0 x 10^-3 m

equivalent capacitance between two capacitors is

C_ek = C₁ + C₂ = ε_rε₀A/d + ε₀A/d = (1 + ε)ε₀A/d

     = (1 + 3.00)(8.85 x 10^-12 C²/Nm²)(5.00 x 10^-3 m²)/(2.0 x 10^-3 m)

C_ek = 8.85 x 10^-11 F = 88.5 pF

Because the arrangement of capacitors is parallel then the voltage at the ends of the capacitors is the same

V₁ = V₂ = V, then

the electrical charge stored on each capacitors is

q₁ = C₁V = [ε_rε₀A/d]V

     = [3.00(8.85 x 10^-12 C²/Nm²)(5.00 x 10^-3 m²)/(2.0 x 10^-3 m)]12.0 V

q₁ = 7.965 x 10^-10 C = 0.8 nC

and

q₂ = q₁/ε_r = 2.655 x 10^-10 C = 0.27 nC

Problem #2
Figure 02 shows a parallel plate capacitor with a plate area A = 5.56 cm² and separation d = 5.56 mm. The left half of the gap is filled with material of dielectric constant k₁ = 7.00; the right half is filled with material of dielectric constant k₂ = 12.0. What is the capacitance?

Fig.2

Answer:
Known:
Plate area A = 5.56 cm² = 5.56 x 10^─4 m²
Separation d = 5.56 mm = 5.56 x 10^─3 m
dielectric constant k₁ = 7.00
dielectric constant k₂ = 12.0

equivalent capacitance between two capacitors is

C_ek = C₁ + C₂ = [k₁ε₀A₁/d₁] + [k₂ε₀A₂/d₂]

C_ek = [k₁ε₀A/2d] + [k₂ε₀A/2d]

C_ek = (ε₀A/2d)(k₁ + k₂)

     = [(8.85 x 10^-12 C²/Nm²)(5.56 x 10^─4 m²)/(5.56 x 10^─3 m)](7.00 + 12.00)

C_ek = 1.68 x 10^─11 F = 16.8 pF

Problem #3
Figure 03 shows a parallel-plate capacitor with a plate area A = 7.89 cm² and plate separation d = 4.62 mm. The top half of the gap is filled with material of dielectric constant k₁ = 11.0; the bottom half is filled with material of dielectric constant k₂ = 12.0. What is the capacitance?

Fig.3

Answer:
Known:
Plate area A = 7.89 cm² = 7.89 x 10^─4 m²
Separation d = 4.62 mm = 4.62 x 10^─3 m
dielectric constant k₁ = 11.0
dielectric constant k₂ = 12.0

the capacitance of each capacitor is

C₁ = k₁ε₀A₁/d₁ = 2 x  (11.0)(8.85 x 10^-12 C²/Nm²)(7.89 x 10^─4 m²)/(4.62 x 10^─3 m) = 3.325 x 10^─11 F = 33.25 pF

C₂ = k₂ε₀A₂/d₂ = 2 x (12.0)(8.85 x 10^-12 C²/Nm²)(7.89 x 10^─4 m²)/(4.62 x 10^─3 m) = 3.627 x 10^─11 F = 36.27 pF

equivalent capacitance between two capacitors is

C_ek = C₁C₂ /(C₁ + C₂)

      = (33.25 pF)(36.27 pF)/(33.25 pF + 36.27 pF)

C_ek = 17.35 pF

Problem #4
Figure 04 shows a parallel plate capacitor of plate area A = 10.5 cm² and plate separation 2d = 7.12 mm. The left half of the gap is filled with material of dielectric constant k₁ = 21.0; the top of the right half is filled with material of dielectric constant k₂ = 42.0; the bottom of the right half is filled with material of dielectric constant k₃ = 58.0. What is the capacitance?

Fig.4

Answer:

Known:
Plate area A = 10.5 cm² = 10.5 x 10^─4 m²
Separation d = 7.12 mm/2 = 3.56 x 10^─3 m
dielectric constant k₁ = 21.0
dielectric constant k₂ = 42.0
dielectric constant k₃ = 58.0

the capacitance of each capacitor is

C₁ = k₁ε₀A₁/d₁ = (21.0)(8.85 x 10^-12 C²/Nm²)(5.25 x 10^─4 m²)/(7.12 x 10^─3 m) = 1.37 x 10^─11 F = 13.7 pF

C₂ = k₂ε₀A₂/d₂ = (42.0)(8.85 x 10^-12 C²/Nm²)(5.25 x 10^─4 m²)/(3.56 x 10^─3 m) = 5.48 x 10^─11 F = 54.8 pF

C₃ = k₃ε₀A₃/d₃ = (56.0)(8.85 x 10^-12 C²/Nm²)(5.25 x 10^─4 m²)/(3.56 x 10^─3 m) = 7.31 x 10^─11 F = 73.1 pF

C₂ and C₃ connected in series

C₂₃ = C₂C₃/(C₂ + C₃) = (54.8 pF)(73.1 pF)/(54.8 pF + 73.1 pF) = 31.32 pF

Equivalent capacitance between two capacitors is (with C₂₃ and C₁ connected in parallel)

C_ek = C₂₃ + C₁ = 31.32 pF + 13.7 pF = 45.0 pF  

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