An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.70 mm.
(a) If a 16.0 V potential difference is applied to these plates, calculate the electric field between the plates.
kV/m
(b) What is the surface charge density?
nC/m2
(c) What is the capacitance?
pF
(d) Find the charge on each plate.
pC
To answer these questions, we can use the formulas related to capacitors and electric fields. Here's how you can calculate each value:
(a) The electric field between the plates can be calculated using the formula:
Electric Field (E) = Voltage (V) / Distance (d)
Given:
Voltage (V) = 16.0 V
Distance (d) = 1.70 mm = 0.0017 m
Substitute the values into the formula to calculate the electric field:
E = 16.0 V / 0.0017 m = 9411.76 V/m
Therefore, the electric field between the plates is 9411.76 V/m.
(b) The surface charge density can be calculated using the formula:
Surface Charge Density (σ) = Electric Field (E) * Permittivity of Free Space (ε0)
The permittivity of free space is a constant value: ε0 = 8.85 × 10⁻¹² F/m
Substitute the value of the electric field to calculate surface charge density:
σ = 9411.76 V/m * 8.85 × 10⁻¹² F/m = 8.33 × 10⁻⁸ C/m²
Therefore, the surface charge density is 8.33 × 10⁻⁸ C/m².
(c) The capacitance can be calculated using the formula:
Capacitance (C) = (Permittivity of Free Space (ε0) x Area (A)) / Distance (d)
Given:
Area (A) = 7.60 cm² = 7.60 × 10⁻⁴ m²
Distance (d) = 1.70 mm = 0.0017 m
Substitute the values into the formula to calculate the capacitance:
C = (8.85 × 10⁻¹² F/m) * (7.60 × 10⁻⁴ m²) / 0.0017 m = 3.92 × 10⁻¹⁰ F
Therefore, the capacitance is 3.92 × 10⁻¹⁰ F, or 3.92 pF.
(d) The charge on each plate can be calculated using the formula:
Charge (Q) = Capacitance (C) * Voltage (V)
Given:
Capacitance (C) = 3.92 × 10⁻¹⁰ F
Voltage (V) = 16.0 V
Substitute the values into the formula to calculate the charge on each plate:
Q = (3.92 × 10⁻¹⁰ F) * (16.0 V) = 6.27 × 10⁻⁹ C
Therefore, the charge on each plate is 6.27 × 10⁻⁹ C, or 6.27 pC.