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 need to use the formulas and principles related to capacitors and electric fields. Let's break down the problem step by step:
(a) Electric field between the plates:
The electric field between the plates of a capacitor is given by 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 values:
V = 16.0 V
d = 1.70 mm = 0.0017 m
Plugging in these values into the formula, we have:
E = 16.0 V / 0.0017 m
Calculating this expression, we find that the electric field between the plates is approximately 9411.8 V/m.
(b) Surface charge density:
The surface charge density of a capacitor is given by the formula:
σ = Q / A
Where σ is the surface charge density, Q is the charge on the plates, and A is the area of each plate.
Given values:
A = 7.60 cm² = 0.00076 m²
To find the surface charge density, we need to determine the charge on the plates. We will find this in part (d) and then use the value to calculate the surface charge density in this part.
(c) Capacitance:
The capacitance of a parallel plate capacitor is given by the formula:
C = ε₀ * (A / d)
Where C is the capacitance, ε₀ is the vacuum permittivity (≈ 8.85 x 10^-12 F/m), A is the area of each plate, and d is the distance between the plates.
Given values:
A = 7.60 cm² = 0.00076 m²
d = 1.70 mm = 0.0017 m
Plugging in these values into the formula, we have:
C = (8.85 x 10^-12 F/m) * (0.00076 m² / 0.0017 m)
Calculating this expression, we find that the capacitance is approximately 3.99 pF.
(d) Charge on each plate:
To find the charge on each plate, we will use the formula:
Q = C * V
Where Q is the charge on each plate, C is the capacitance, and V is the potential difference applied to the plates.
Given values:
C = 3.99 pF = 3.99 x 10^-12 F
V = 16.0 V
Plugging in these values into the formula, we have:
Q = (3.99 x 10^-12 F) * (16.0 V)
Calculating this expression, we find that the charge on each plate is approximately 6.38 pC.
So, to summarize:
(a) The electric field between the plates is approximately 9411.8 V/m.
(b) The surface charge density is dependent on the charge on the plates, which we will calculate in part (d).
(c) The capacitance is approximately 3.99 pF.
(d) The charge on each plate is approximately 6.38 pC.