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 formulas related to capacitors and electric fields.

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

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 x 10^(-3) m,

Plugging in these values, we get:

E = 16.0 V / (1.70 x 10^(-3) m) = 9411 V/m.

The electric field between the plates is 9411 V/m.

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

σ = ε0 * E,

where σ is the surface charge density, ε0 is the permittivity of free space, and E is the electric field.

Given:
ε0 = 8.85 x 10^(-12) F/m (permittivity of free space),
E = 9411 V/m.

Plugging in these values, we get:

σ = (8.85 x 10^(-12) F/m) * (9411 V/m) = 8.31 x 10^(-8) C/m^2.

The surface charge density is 8.31 x 10^(-8) C/m^2.

(c) The capacitance of the capacitor can be calculated using the formula:

C = (ε0 * A) / d,

where C is the capacitance, A is the area of one plate, and d is the distance between the plates.

Given:
A = 7.60 cm^2 = 7.60 x 10^(-4) m^2,
d = 1.70 mm = 1.70 x 10^(-3) m,
ε0 = 8.85 x 10^(-12) F/m.

Plugging in these values, we find:

C = (8.85 x 10^(-12) F/m) * (7.60 x 10^(-4) m^2) / (1.70 x 10^(-3) m) = 3.94 x 10^(-14) F.

The capacitance is 3.94 x 10^(-14) F, or 39.4 pF.

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

Q = C * V,

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

Given:
C = 3.94 x 10^(-14) F,
V = 16.0 V.

Plugging in these values, we find:

Q = (3.94 x 10^(-14) F) * (16.0 V) = 6.30 x 10^(-13) C.

The charge on each plate is 6.30 x 10^(-13) C, or 630 pC.

To answer these questions, we will need to use the formulas related to capacitors and electric fields.

(a) The electric field between the plates of a capacitor with a potential difference can be determined 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 = 0.00170 m

E = 16.0 V / 0.00170 m = 9411.76 V/m

Therefore, the electric field between the plates is 9411.76 V/m.

(b) The surface charge density on the plates can be determined using the formula:

σ = q / A

where σ is the surface charge density, q is the charge on the plate, and A is the area of the plate.

Since the plates of a capacitor have equal magnitudes of charge but opposite signs, we can consider only one plate. Let's assume the charge on each plate is q.

Plugging in the given values:
A = 7.60 cm^2 = 7.60 × 10^(-4) m^2

σ = q / 7.60 × 10^(-4) m^2

We need more information to find the charge on each plate in part (d) in order to calculate the surface charge density.

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

C = ε₀ * (A / d)

where C is the capacitance, ε₀ is the permittivity of free space (ε₀ = 8.85 × 10^(-12) F/m), A is the area of the plate, and d is the distance between the plates.

Plugging in the given values:
A = 7.60 cm^2 = 7.60 × 10^(-4) m^2
d = 1.70 mm = 0.00170 m

C = (8.85 × 10^(-12) F/m) * (7.60 × 10^(-4) m^2 / 0.00170 m)
= 3.95 × 10^(-11) F
= 39.5 pF

Therefore, the capacitance of the capacitor is 39.5 pF.

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

Q = C * V

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

Plugging in the given values:
C = 39.5 pF = 39.5 × 10^(-12) F
V = 16.0 V

Q = (39.5 × 10^(-12) F) * (16.0 V)
= 6.32 × 10^(-10) C
= 632 pC

Therefore, the charge on each plate is 632 pC.