Question:

A parallel-plate capacitor consists of two plates with an area of 0.02 m² each, separated by a distance of 0.005 m. The plates are made of a material with a permittivity of vacuum of ε₀ = 8.85 × 10⁻¹² F/m. The capacitor is connected to a 12 V battery.

a) Calculate the capacitance of the capacitor. b) Determine the charge stored on each plate of the capacitor. c) Find the electric field magnitude between the plates. d) Determine the energy stored in the capacitor.

Answer:

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

C = (ε₀ * A) / d

Where C is the capacitance, ε₀ is the permittivity of vacuum, A is the area of each plate, and d is the separation between the plates.

Given: A = 0.02 m² d = 0.005 m ε₀ = 8.85 × 10⁻¹² F/m

Substituting the given values into the formula, we can calculate the capacitance:

C = (8.85 × 10⁻¹² F/m * 0.02 m²) / 0.005 m C = 35.4 × 10⁻¹² F

Therefore, the capacitance of the capacitor is 35.4 picofarads (pF).

b) The charge stored on each plate of the capacitor can be determined using the formula:

Q = C * V

Where Q is the charge, C is the capacitance, and V is the voltage across the capacitor.

Given: C = 35.4 × 10⁻¹² F V = 12 V

Substituting the given values into the formula, we can calculate the charge on each plate:

Q = (35.4 × 10⁻¹² F) * 12 V Q = 424.8 × 10⁻¹² C

Therefore, the charge stored on each plate of the capacitor is 424.8 picocoulombs (pC).

c) The electric field magnitude between the plates of a parallel-plate capacitor can be calculated using the formula:

E = V / d

Where E is the electric field magnitude, V is the voltage across the capacitor, and d is the separation between the plates.

Given: V = 12 V d = 0.005 m

Substituting the given values into the formula, we can calculate the electric field magnitude:

E = 12 V / 0.005 m E = 2400 V/m

Therefore, the electric field magnitude between the plates of the capacitor is 2400 volts per meter (V/m).

d) The energy stored in a capacitor can be determined using the formula:

U = (1/2) * C * V²

Where U is the energy, C is the capacitance, and V is the voltage across the capacitor.

Given: C = 35.4 × 10⁻¹² F V = 12 V

Substituting the given values into the formula, we can calculate the energy stored in the capacitor:

U = (1/2) * (35.4 × 10⁻¹² F) * (12 V)² U = 3.211 × 10⁻¹⁰ J

Therefore, the energy stored in the capacitor is 3.211 × 10⁻¹⁰ joules (J).