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 this question, we can use the formulas and equations related to capacitance. Let's go step-by-step:

(a) The electric field between the plates can be calculated using the formula:
E = V/d
where E is the electric field, V is the potential difference, and d is the distance between the plates.

Plugging in the given values:
V = 16.0 V
d = 1.70 mm = 0.0017 m

E = 16.0 V / 0.0017 m
E ≈ 9411 kV/m

Therefore, the electric field between the plates is approximately 9411 kV/m.

(b) The surface charge density can be calculated using the formula:
σ = ε0 * E
where σ is the surface charge density, ε0 is the permittivity of free space (8.85 x 10^(-12) F/m), and E is the electric field.

Plugging in the given value:
ε0 = 8.85 x 10^(-12) F/m
E = 9411 kV/m = 9411 x 10^3 V/m

σ = (8.85 x 10^(-12) F/m) * (9411 x 10^3 V/m)
σ ≈ 8.32 nC/m^2

Therefore, the surface charge density is approximately 8.32 nC/m^2.

(c) The capacitance can be calculated using the formula:
C = ε0 * A / d
where C is the capacitance, ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

Plugging in the given values:
ε0 = 8.85 x 10^(-12) F/m
A = 7.60 cm^2 = 7.60 x 10^(-4) m^2
d = 1.70 mm = 0.0017 m

C = (8.85 x 10^(-12) F/m) * (7.60 x 10^(-4) m^2) / 0.0017 m
C ≈ 3.92 pF

Therefore, the capacitance is approximately 3.92 pF.

(d) The charge on each plate can be calculated using the formula:
Q = C * V
where Q is the charge, C is the capacitance, and V is the voltage (potential difference).

Plugging in the given values:
C = 3.92 pF = 3.92 x 10^(-12) F
V = 16.0 V

Q = (3.92 x 10^(-12) F) * (16.0 V)
Q ≈ 6.27 pC

Therefore, the charge on each plate is approximately 6.27 pC.

To answer these questions, we can make use of the formulas and principles related to capacitance, electric field, and surface charge density. Let's go through each question step by step:

(a) To calculate the electric field between the plates, we can use the formula for electric field intensity between two parallel plates:

E = V/d,

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

Substituting the given values, we have:

E = 16.0 V / (1.70 mm * 10^-3 m/mm).

Calculating this expression, we get:

E ≈ 9400 V/m.

Therefore, the electric field between the plates is 9400 V/m.

(b) The surface charge density (σ) is defined as the amount of charge (Q) per unit area (A) on the capacitor plates. Mathematically, we have:

σ = Q / A.

Given the area of each plate (A) as 7.60 cm^2 and assuming the charge on each plate is equal, we can calculate the surface charge density by dividing twice the charge on one plate by the total area of both plates:

σ = 2Q / (2A) = Q / A.

(c) The capacitance (C) of the air-filled capacitor is given by the formula:

C = ε₀ * (A/d),

where ε₀ is the permittivity of free space, and A and d are the same as given in the question.

The permittivity of free space (ε₀) is approximately 8.854 × 10^-12 F/m. Substituting the values, we have:

C = (8.854 × 10^-12 F/m) * (7.60 cm^2 * 10^-4 m^2/cm^2) / (1.70 mm * 10^-3 m/mm).

Calculating this expression, we get:

C ≈ 3.55 pF.

Therefore, the capacitance of the air-filled capacitor is approximately 3.55 pF.

(d) To find the charge on each plate, we can use the formula:

Q = C * V,

where C is the capacitance and V is the potential difference. Substituting the values, we have:

Q = (3.55 pF) * (16.0 V).

Calculating this expression, we have:

Q ≈ 56.8 pC.

Therefore, the charge on each plate is approximately 56.8 pC.