Post

Calculating Capacitance of a Parallel Plate Capacitor

Created by @nathanedwards
 at November 3rd 2023, 5:12:59 am.
AP_Physics_2-Questions
#physics
#capacitance
#parallel-plate-capacitor

Question:

Two parallel plates of a capacitor are separated by a distance of d=0.02d = 0.02 m. The area of each plate is A=0.01A = 0.01 m2. The region between the plates is filled with a dielectric material with a dielectric constant of k=4k = 4. Find the capacitance of this capacitor.

Answer:

The capacitance of a parallel plate capacitor can be calculated using the formula:

C=ε0⋅A⋅kd\large C = \frac{{\varepsilon_0 \cdot A \cdot k}}{{d}}

where:

  • CC is the capacitance,
  • ε0\varepsilon_0 is the vacuum permittivity and has a value of 8.85×10−128.85 \times 10^{-12} F/m,
  • AA is the area of each plate,
  • kk is the dielectric constant, and
  • dd is the separation between the plates.

Substituting the given values into the formula, we have:

C=8.85×10−12 F/m×0.01 m2×40.02 m\large C = \frac{{8.85 \times 10^{-12} \, \text{F/m} \times 0.01 \, \text{m}^2 \times 4}}{{0.02 \, \text{m}}}

Simplifying the expression, we get:

C=8.85×10−12 F/m×0.01 m2×40.02 m=1.77×10−9 F\large C = \frac{{8.85 \times 10^{-12} \, \text{F/m} \times 0.01 \, \text{m}^2 \times 4}}{{0.02 \, \text{m}}} = 1.77 \times 10^{-9} \, \text{F}

Therefore, the capacitance of this capacitor is 1.77×10−91.77 \times 10^{-9} F.

Chat

Save To Bookmark