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 need to use the formula relevant to capacitors:

C = ε₀ * (A / d) [Capacitance formula]
E = V / d [Electric field formula]
σ = Q / A [Surface charge density formula]

where:
C is the capacitance,
ε₀ is the vacuum permittivity (8.85 x 10^-12 F/m),
A is the area of the plates,
d is the distance between the plates,
E is the electric field,
V is the potential difference,
σ is the surface charge density, and
Q is the charge on the plates.

Now, let's calculate the solutions step by step:

(a) Electric field (E):
Using E = V / d, substitute V = 16.0 V and d = 1.70 mm (or 1.70 x 10^-3 m) into the equation:
E = 16.0 V / (1.70 x 10^-3 m)

Calculate the result to find the value of electric field E in kV/m.

(b) Surface charge density (σ):
Using σ = Q / A, we need to find the charge on the plates (Q) first. Once we have Q, we can use the formula σ = Q / A to find the surface charge density.

(c) Capacitance (C):
Using the capacitance formula, C = ε₀ * (A / d), substitute the values of ε₀, A, and d into the equation:
C = (8.85 x 10^-12 F/m) * (7.60 cm^2 / (1.70 x 10^-3 m))

Calculate the result to find the value of capacitance C in pF.

(d) Charge on each plate (Q):
To find the charge on each plate, we can use either Q = CV or σ = Q / A.
Since we already have the value of C from part (c), we can use Q = CV, where Q = charge on each plate, C = capacitance, and V = potential difference.
Q = (capacitance C) * (potential difference V)

Finally, calculate the result to find the charge on each plate Q in pC.