AP Physics 2 Exam Question:

A parallel plate capacitor is made by sandwiching a dielectric material between two square plates. The plates have a side length of 10 cm and are separated by a distance of 1.5 mm. The dielectric material has a relative permittivity of 2.5. A potential difference of 12 V is applied across the capacitor.

a) Calculate the capacitance of the capacitor.

b) If the potential difference across the capacitor is increased to 20 V while the distance between the plates and the dielectric material remain unchanged, calculate the new electric field between the plates.

c) Determine the electric potential energy stored in the capacitor both before and after the potential difference was increased, if the charge stored in the capacitor remains unchanged.

Answer:

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

C = (ε₀ * εᵣ * A) / d

where C is the capacitance, ε₀ is the vacuum permittivity (8.85 x 10⁻¹² F/m), εᵣ is the relative permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates.

Given values: A = (10 cm)² = 0.01 m², d = 1.5 mm = 0.0015 m, and εᵣ = 2.5.

Substituting these values into the equation, we can calculate the capacitance:

C = (8.85 x 10⁻¹² F/m * 2.5 * 0.01 m²) / 0.0015 m
C ≈ 1.177 F

Therefore, the capacitance of the capacitor is approximately 1.177 F.

b) To calculate the new electric field between the plates, we can use the formula:

E = V / d

where E is the electric field, V is the potential difference, and d is the distance between the plates.

Given values: V = 20 V and d = 0.0015 m.

Substituting these values into the equation, we can calculate the new electric field:

E = 20 V / 0.0015 m
E ≈ 13333.33 V/m

Therefore, the new electric field between the plates is approximately 13333.33 V/m.

c) The electric potential energy stored in a capacitor is given by the formula:

U = (1/2) * C * V²

where U is the electric potential energy, C is the capacitance, and V is the potential difference.

Before the potential difference was increased, the values were: C = 1.177 F and V = 12 V.

Substituting these values into the equation, we can calculate the initial electric potential energy:

U₀ = (1/2) * 1.177 F * (12 V)²
U₀ ≈ 84.58 J

Therefore, the electric potential energy stored in the capacitor before the potential difference was increased is approximately 84.58 J.

After the potential difference was increased, the values are: C = 1.177 F and V = 20 V.

Substituting these values into the equation, we can calculate the new electric potential energy:

U = (1/2) * 1.177 F * (20 V)²
U ≈ 235.4 J

Therefore, the electric potential energy stored in the capacitor after the potential difference was increased, if the charge stored in the capacitor remains unchanged, is approximately 235.4 J.