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Parallel-Plate Capacitor Properties

Created by @nathanedwards
 at November 1st 2023, 6:28:15 pm.
AP_Physics_2-Questions
#physics
#electric-field
#capacitance

Question:

A parallel-plate capacitor is formed by two square plates separated by a distance of 2 cm. The plates have lengths of 10 cm on each side. The space between the plates is filled with a dielectric material that has a relative permittivity of 5.

a) Determine the capacitance of this capacitor.

b) If a potential difference of 100 V is applied across the plates, calculate the charge stored on each plate.

c) If the separation between the plates is now increased to 5 cm while keeping the area of each plate constant, calculate the new capacitance.

d) If the electric field between the plates is 6 N/C, calculate the magnitude of the surface charge density on each plate.

Assume the electric field between the plates is uniform and neglect fringing effects.

Answer:

a) The capacitance CC of a parallel-plate capacitor is given by the formula:

C=ε0εrAd C = \frac{{\varepsilon_0 \varepsilon_r A}}{d}

where:

  • ε0\varepsilon_0 is the permittivity of free space (8.85×10−12 F/m8.85 \times 10^{-12} \, \text{F/m})
  • εr\varepsilon_r is the relative permittivity of the dielectric material
  • AA is the area of each plate
  • dd is the separation between the plates

Plugging in the known values:

C=(8.85×10−12 F/m)(5)(10−2 m)22×10−2 m C = \frac{{(8.85 \times 10^{-12} \, \text{F/m})(5)(10^{-2} \, \text{m})^2}}{2 \times 10^{-2} \, \text{m}}

Calculating the capacitance:

C=2.21×10−10 F C = 2.21 \times 10^{-10} \, \text{F}

b) The charge QQ stored on each plate of a capacitor is given by the formula:

Q=CV Q = CV

where:

  • VV is the potential difference across the plates

Plugging in the known values:

Q=(2.21×10−10 F)(100 V) Q = (2.21 \times 10^{-10} \, \text{F})(100 \, \text{V})

Calculating the charge:

Q=2.21×10−8 C Q = 2.21 \times 10^{-8} \, \text{C}

c) When the separation between the plates is increased to 5 cm while keeping the area constant, the new capacitance C′C' can be calculated using the formula from part a:

C′=(8.85×10−12 F/m)(5)(10−2 m)25×10−2 m C' = \frac{{(8.85 \times 10^{-12} \, \text{F/m})(5)(10^{-2} \, \text{m})^2}}{5 \times 10^{-2} \, \text{m}}

Calculating the new capacitance:

C′=4.43×10−10 F C' = 4.43 \times 10^{-10} \, \text{F}

d) The electric field EE between the plates of a parallel-plate capacitor is given by the formula:

E=Vd E = \frac{V}{d}

where:

  • VV is the potential difference across the plates
  • dd is the separation between the plates

Plugging in the known values:

6 N/C=V2×10−2 m 6 \, \text{N/C} = \frac{V}{2 \times 10^{-2} \, \text{m}}

Solving for VV:

V=1.2 V V = 1.2 \, \text{V}

The surface charge density σ\sigma on each plate of a capacitor is given by the formula:

σ=QA \sigma = \frac{Q}{A}

where:

  • QQ is the charge stored on each plate
  • AA is the area of each plate

Plugging in the known values:

σ=2.21×10−8 C(10−2 m)2 \sigma = \frac{2.21 \times 10^{-8} \, \text{C}}{(10^{-2} \, \text{m})^2}

Calculating the surface charge density:

σ=2.21×10−4 C/m2 \sigma = 2.21 \times 10^{-4} \, \text{C/m}^2

Therefore, the magnitude of the surface charge density on each plate is 2.21×10−4 C/m22.21 \times 10^{-4} \, \text{C/m}^2.

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