AP Physics 2 Exam Question: Capacitance

A parallel plate capacitor consists of two square plates, each with side length 's' and separation distance 'd'. The space between the plates is filled with a dielectric material of constant κ (kappa).

1. Determine the formula for the capacitance 'C' of this parallel plate capacitor.

2. A capacitor is constructed with square plates of side length 0.05 m and separation distance 0.01 m. The dielectric material between the plates has a constant κ = 3. Find the capacitance of this capacitor.

Answer:

1. The formula for the capacitance 'C' of a parallel plate capacitor is given by the equation:

   

   where:

   • C is the capacitance (in Farads, F),
   • ε₀ is the vacuum permittivity (8.85 x 10⁻¹² F/m),
   • εᵣ is the relative permittivity (or dielectric constant) of the material between the plates,
   • A is the area of each plate (in square meters, m²), and
   • d is the separation distance between the plates (in meters, m).

2. Given:

   • Side length of plates (s) = 0.05 m
   • Separation distance between plates (d) = 0.01 m
   • Dielectric constant (κ) = 3

   We can calculate the area of each plate using the formula A = s²:

   

   Substituting the given values, we find:

   

   Now, substituting the values into the capacitance formula:

   

   Evaluating the above expression, we get:

   

   Therefore, the capacitance of the given capacitor is approximately 2.11 x 10⁻¹⁰ F.

Note: The value of capacitance obtained can be rounded to an appropriate number of significant figures as required by the specific problem or exam prompt.