Uningenia
Start studyingStart
ElectrostaticsHard

Parallel-plate capacitor with dielectric

A parallel-plate capacitor has plates of area A = 400 cm² and separation d = 2 mm. It is filled with a dielectric of constant εᵣ = 5 and connected to V = 100 V.\nCalculate: (a) capacitance, (b) charge on the plates, (c) internal electric field E, (d) stored energy.
Given data
A = 400×10⁻⁴ m² = 0.04 m²d = 2×10⁻³ mεᵣ = 5V = 100 Vε₀ = 8.85×10⁻¹² F/m
Review the theory: Elettrostatica
Steps
0 / 4
  1. Calculate the capacitance C with the dielectric (in pF = 10⁻¹² F).
  2. Calculate the charge Q on the plates (in nC = 10⁻⁹ C).
  3. Calculate the electric field E inside the capacitor (in V/m).
  4. Calculate the energy U stored in the capacitor (in μJ = 10⁻⁶ J).
Full worked solution
  1. Calculate the capacitance C with the dielectric (in pF = 10⁻¹² F).
    C=εrε0Ad=5⋅8.85×10−12⋅0.040.002C = \varepsilon_r \varepsilon_0 \dfrac{A}{d} = 5 \cdot 8.85\times10^{-12} \cdot \dfrac{0.04}{0.002}
    C=εrε0Ad=5×8.85×10−12×0.040.002=5×8.85×10−12×20=885×10−12 F=885 pFC = \varepsilon_r \varepsilon_0 \frac{A}{d} = 5 \times 8.85\times10^{-12} \times \frac{0.04}{0.002} = 5 \times 8.85\times10^{-12} \times 20 = \mathbf{885\times10^{-12}\,F} = \mathbf{885\,pF}. Without the dielectric, the capacitance would be 177 pF177\,\mathrm{pF} (5 times smaller).
  2. Calculate the charge Q on the plates (in nC = 10⁻⁹ C).
    Q=C⋅V=885×10−12⋅100Q = C \cdot V = 885\times10^{-12} \cdot 100
    Q=C V=885×10−12×100=8.85×10−8 C=88.5 nCQ = C\,V = 885\times10^{-12} \times 100 = \mathbf{8.85\times10^{-8}\,C} = \mathbf{88.5\,nC}. The charge stored on the plates is the product of capacitance and voltage.
  3. Calculate the electric field E inside the capacitor (in V/m).
    E=Vd=1000.002E = \dfrac{V}{d} = \dfrac{100}{0.002}
    E=V/d=100/0.002=50000 V/m=50 kV/mE = V/d = 100 / 0.002 = \mathbf{50000\,V/m} = \mathbf{50\,kV/m}. The electric field inside a parallel-plate capacitor depends only on VV and dd, not on the dielectric.
  4. Calculate the energy U stored in the capacitor (in μJ = 10⁻⁶ J).
    U=12CV2=12⋅885×10−12⋅1002U = \dfrac{1}{2}CV^2 = \dfrac{1}{2} \cdot 885\times10^{-12} \cdot 100^2
    U=12CV2=12×885×10−12×1002=12×885×10−12×10000=4.425×10−6 J=4.43 μJU = \frac{1}{2}C V^2 = \frac{1}{2} \times 885\times10^{-12} \times 100^2 = \frac{1}{2} \times 885\times10^{-12} \times 10000 = \mathbf{4.425\times10^{-6}\,J} = \mathbf{4.43\,\mu J}.
Result:C = 885 pF — Q = 88.5 nC — E = 50 kV/m — U = 4.43 μJ.
TheoryElettrostaticaFormula sheetPhysics 2

Similar exercises