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

(a) To calculate the electric field between the plates, we can use the formula for the electric field created by a capacitor:

E = V / d

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

Given:
V = 16.0 V
d = 1.70 mm = 1.70 × 10^(-3) m

Plugging in these values into the formula, we get:

E = 16.0 V / (1.70 × 10^(-3) m)
E ≈ 9411.8 V/m (or 9.41 kV/m)

Therefore, the electric field between the plates is approximately 9.41 kV/m.

(b) Surface charge density (σ) can be calculated using the formula:

σ = ε₀ * E

Where ε₀ is the permittivity of free space and E is the electric field.

Given:
E = 9.41 kV/m = 9.41 × 10^3 V/m

The value of ε₀ is approximately 8.85 × 10^(-12) F/m.

Plugging in these values into the formula, we get:

σ = (8.85 × 10^(-12) F/m) * (9.41 × 10^3 V/m)
σ ≈ 8.3068 × 10^(-8) C/m² (or 83.068 nC/m²)

Therefore, the surface charge density is approximately 83.068 nC/m².

(c) The capacitance (C) of a parallel plate capacitor can be calculated using the formula:

C = (ε₀ * A) / d

Where ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

Given:
A = 7.60 cm² = 7.60 × 10^(-4) m²
d = 1.70 mm = 1.70 × 10^(-3) m

Plugging in these values into the formula, we get:

C = (8.85 × 10^(-12) F/m) * (7.60 × 10^(-4) m²) / (1.70 × 10^(-3) m)
C ≈ 3.9518 × 10^(-11) F (or 39.518 pF)

Therefore, the capacitance is approximately 39.518 pF.

(d) The charge (Q) on each plate can be calculated using the formula:

Q = C * V

Where C is the capacitance and V is the potential difference.

Given:
C = 39.518 pF = 39.518 × 10^(-12) F
V = 16.0 V

Plugging in these values into the formula, we get:

Q = (39.518 × 10^(-12) F) * (16.0 V)
Q ≈ 6.3229 × 10^(-9) C (or 6.3229 pC)

Therefore, the charge on each plate is approximately 6.3229 pC.

To solve these problems, we need to use some equations and formulas related to the electric field, capacitance, and charge of a capacitor. Let's break down each question and go step by step to find the solutions.

(a) To find the electric field between the plates, we can use the formula:

Electric field (E) = Voltage (V) / Distance between the plates (d)

In this case, the voltage is given as 16.0 V, and the distance between the plates is given as 1.70 mm.
Converting 1.70 mm to meters, we get: 1.70 mm = 0.0017 m

Plugging in the values into the formula, we have:

E = 16.0 V / 0.0017 m = 9411 V/m

So, the electric field between the plates is 9411 V/m.

(b) To find the surface charge density, we can use the equation:

Surface charge density (σ) = Electric field (E) * Absolute permittivity (ε₀)

The absolute permittivity, denoted as ε₀, is a constant value equal to 8.85 x 10^-12 F/m.

Plugging in the values, we have:

σ = 9411 V/m * 8.85 x 10^-12 F/m = 8.327 x 10^-8 C/m²

So, the surface charge density is 8.327 x 10^-8 C/m².

(c) To find the capacitance, we can use the formula:

Capacitance (C) = (ε₀ * Area) / Distance between the plates

The area of the plates is given as 7.60 cm², which we convert to meters:
7.60 cm² = 0.00076 m²

Plugging in the values, we have:

C = (8.85 x 10^-12 F/m * 0.00076 m²) / 0.0017 m
C = 3.964 x 10^-14 F
Since the capacitance is often represented in picofarads (pF), we can convert it:

1 F = 1 x 10^12 pF
So, 3.964 x 10^-14 F = 3.964 x 10^-2 pF

Therefore, the capacitance is 3.964 x 10^-2 pF.

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

Charge (Q) = Capacitance (C) * Voltage (V)

Plugging in the values, we have:

Q = 3.964 x 10^-2 pF * 16.0 V
Q = 6.343 x 10^-1 pC

Therefore, the charge on each plate is 6.343 x 10^-1 pC.

In summary:
(a) The electric field between the plates is 9411 V/m.
(b) The surface charge density is 8.327 x 10^-8 C/m².
(c) The capacitance is 3.964 x 10^-2 pF.
(d) The charge on each plate is 6.343 x 10^-1 pC.