Question

A dielectric-filled parallel-plate capacitor has plate area A = 25.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 2.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2

1. Find the energy U1 of the dielectric-filled capacitor.

2. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

3. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.

4. In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric? Express your answers in joules

Answer 1

A = (25 cm²) ⁄ (100 cm ⁄ m)² = 0.0025 m²

 d = 0.0010 m

       C = (Єø) • (Єʀ) • A ⁄ d
       C = [8.85 ×10^(-12)] • (2) • (0.0025) ⁄ (0.001)
       C = 4.425 ×10^(-12) F
      
  U = (½) • C • V²
  U = (½) • [8.85 ×10^(-12)] • (15²)
  U = 4.98 ×10^(-10) Joules
 
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 At the halfway point there are essentially two capacitors in parallel

   Cɪ = (Єø)  • A ⁄ d ... (air region)
   Cɪ = [8.85 ×10^(-12)] • (0.00125) ⁄ (0.001)
   Cɪ = 1.106 ×10^(-12) F
   Cɪ = 1.106 pF

       Cɪɪ = (Єø) • (Єʀ) • A ⁄ d ... (dielectric region)
       Cɪɪ = [8.85 ×10^(-12)] • (2) • (0.00125) ⁄ (0.001)
       Cɪɪ = 2.212 ×10^(-12) F
       Cɪɪ = 2.212 pF

  Cт = Cɪ + Cɪɪ
  Cт = 1.106 + 2.212 = 3.318 pF

     U_2 = (½) • C • V²
     U_2 = (½) • [3.318 ×10^(-12)] • (15²)
     U_2 = 3.735 ×10^(-10) Joules
     
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  Q is conserved in this case:

      Q = C • V
      Q = [3.32 ×10^(-12)] • (15)
      Q = 4.98 ×10^(-11) C
     


   C = (Єø)  • A ⁄ d ... (completely air)
   C = [8.85 ×10^(-12)] • (0.0025) ⁄ (0.001)
   C = 2.21 ×10^(-12) F
   C = 2.21 pF

     U_3 = (½) • C • V²
     U_3 = (½) • C • (Q ⁄ C)²
     U_3 = (½) • Q² ⁄ C
     U_3 = (½) • [4.98 ×10^(-11)]² ⁄ 2.21×10^(-12)
     U_3 = 5.61 X 10^-10J
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  Work = ∆U = ( 5.61 X 10^-10) − (37.3 x 10^-9J)

  Work = 1.875 X 10^-10 J