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 solve this problem, we can use the following equations:

(a) Electric field between the plates of a capacitor is given by:

E = V / d

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

(b) Surface charge density is the charge per unit area on the plates, given by:

σ = Q / A

where σ is the surface charge density, Q is the charge on each plate, and A is the area of the plates.

(c) The capacitance of a capacitor is given by:

C = Q / V

where C is the capacitance, Q is the charge, and V is the potential difference.

(d) The charge on each plate is given by:

Q = C * V

where Q is the charge, C is the capacitance, and V is the potential difference.

Now, let's solve each part of the problem step by step:

(a) Electric field between the plates:

E = V / d
E = 16.0 V / (1.70 mm * 10^-3 / 1 m)
E = 16.0 V / 0.0017 m
E = 9411.76 V/m
E ≈ 9.41 kV/m

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

(b) Surface charge density:

σ = Q / A
σ = (Q / V) / A
σ = (C * V / V) / A
σ = C / A
σ = C / (7.60 cm^2 * (1 m / 100 cm)^2)
σ = C / (7.60 * 10^-4 m^2)
σ = C / 0.00076 m^2

(c) Capacitance:

C = Q / V
C = (Q / A) / V
C = σ / V
C = (Q / (7.60 cm^2 * (1 m / 100 cm)^2)) / 16.0 V
C = Q / (7.60 * 10^-4 m^2 * 16.0 V)
C = Q / 0.01216 m^2V
C = Q / (12.16 * 10^-6 m^2V)

(d) Charge on each plate:

Q = C * V
Q = (Q / (12.16 * 10^-6 m^2V)) * 16.0 V
Q = Q / (0.019456 m^2)
Q * 0.019456 m^2 = Q
Q = 0.019456 Coulombs

Since the unit prefix for charge is pC (picocoulombs), the charge on each plate is approximately 19.46 pC.

So, the answers to the given questions are:

(a) The electric field between the plates is approximately 9.41 kV/m.
(b) The surface charge density is C / 0.00076 m^2.
(c) The capacitance is Q / (12.16 * 10^-6 m^2V).
(d) The charge on each plate is approximately 19.46 pC.

To answer these questions, we can use the formulas and equations related to capacitors.

(a) The electric field between the plates of a capacitor 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.

In this case, the potential difference is given as 16.0 V, and the distance between the plates is 1.70 mm. First, let's convert the distance to meters:
d = 1.70 mm = 1.70 × 10^-3 m

We can now substitute the values into the formula:
E = 16.0 V / (1.70 × 10^-3 m)

Calculating this will give us the electric field between the plates, expressed in kV/m.

(b) The surface charge density of a capacitor is the charge per unit area on one plate. It can be calculated using the formula σ = Q / A, where σ is the surface charge density, Q is the charge on one plate, and A is the area of one plate.

In this case, the area is given as 7.60 cm^2. First, let's convert the area to square meters:
A = 7.60 cm^2 = 7.60 × 10^-4 m^2

To find the surface charge density, we need to calculate the charge on one plate.

(c) The capacitance of a capacitor 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 one plate, and d is the distance between the plates.

In this case, we are given the area as 7.60 cm^2 and the distance as 1.70 mm. We already converted the area to square meters in part (b). We can now substitute the values into the formula:
C = ε₀ * (7.60 × 10^-4 m^2 / 1.70 × 10^-3 m)

Calculating this will give us the capacitance, expressed in picofarads (pF).

(d) The charge on each plate of a capacitor is equal in magnitude and opposite in sign. It can be calculated using the formula Q = C * V, where Q is the charge, C is the capacitance, and V is the potential difference.

In this case, we are given the capacitance as calculated in part (c), and the potential difference is given as 16.0 V. We can substitute the values into the formula:
Q = (capacitance in pF) * (16.0 V)

Calculating this will give us the charge on each plate, expressed in picocoulombs (pC).