A large capacitance of 1.40 mF is needed for a certain application. (a) Calculate the area the parallel plates of such a capacitor must have if they are separated by 2.00 µm of Teflon. m2 (b) What is the maximum voltage that can be applied? V (c) Find the maximum charge that can be stored. C (d) Calculate the volume of Teflon alone in the capacitor. m3
(a) The capacitance of a parallel-plate capacitor is given by the formula:
C = ε₀(A/d)
where C is the capacitance, ε₀ is the permittivity of free space (ε₀ = 8.85 x 10⁻¹² F/m), A is the area of the plates, and d is the separation between the plates.
Rearranging the formula, we can find the area A:
A = Cd/ε₀
Plugging in the given values:
C = 1.40 mF = 1.40 x 10⁻³ F
d = 2.00 µm = 2.00 x 10⁻⁶ m
ε₀ = 8.85 x 10⁻¹² F/m
A = (1.40 x 10⁻³ F)(2.00 x 10⁻⁶ m)/(8.85 x 10⁻¹² F/m)
= 3.16 x 10⁻⁷ m² (rounded to two significant figures)
Therefore, the area of the parallel plates must be 3.16 x 10⁻⁷ m².
(b) The maximum voltage that can be applied across a capacitor is given by the formula:
V = Q/C
where V is the voltage, Q is the charge stored in the capacitor, and C is the capacitance.
Rearranging the formula, we can find the maximum voltage V:
V = QC
The maximum charge that can be stored in the capacitor is given by:
Q = CV
Plugging in the given values:
C = 1.40 mF = 1.40 x 10⁻³ F
Q = (1.40 x 10⁻³ F)(V) = 1.40 x 10⁻³V
Now we can solve for V:
V = (1.40 x 10⁻³ F)(V)/(1.40 x 10⁻³ F)
= V
Therefore, the maximum voltage that can be applied is V.
(c) The maximum charge that can be stored in the capacitor is given by:
Q = CV
Plugging in the given values:
C = 1.40 mF = 1.40 x 10⁻³ F
V = V
Q = (1.40 x 10⁻³ F)(V)
= 1.40 x 10⁻³V
Therefore, the maximum charge that can be stored is 1.40 x 10⁻³V C.
(d) The volume of Teflon alone in the capacitor is given by:
V = Ad
where V is the volume, A is the area of the plates, and d is the separation between the plates.
Plugging in the given values:
A = 3.16 x 10⁻⁷ m²
d = 2.00 µm = 2.00 x 10⁻⁶ m
V = (3.16 x 10⁻⁷ m²)(2.00 x 10⁻⁶ m)
= 6.32 x 10⁻¹³ m³ (rounded to two significant figures)
Therefore, the volume of Teflon alone in the capacitor is 6.32 x 10⁻¹³ m³.
To answer these questions, we can use the formula for the capacitance of a parallel-plate capacitor:
C = ε₀ * (A / d)
where C is the capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.
Let's calculate each part step by step:
(a) Calculate the area the parallel plates of such a capacitor must have if they are separated by 2.00 µm of Teflon.
Rearranging the formula:
A = (C * d) / ε₀
Given:
C = 1.40 mF = 1.40 x 10^-3 F
d = 2.00 µm = 2.00 x 10^-6 m
ε₀ = 8.85 x 10^-12 F/m (permittivity of free space)
Substituting the given values:
A = (1.40 x 10^-3 F * 2.00 x 10^-6 m) / (8.85 x 10^-12 F/m)
Simplifying the equation:
A = 3.16 m^2
Therefore, the area of the parallel plates must be 3.16 m².
(b) What is the maximum voltage that can be applied?
The maximum voltage that can be applied is related to the breakdown voltage of Teflon. Assuming there's no other information provided, we'll consider the breakdown voltage of Teflon to be approximately 500 V.
Therefore, the maximum voltage that can be applied is 500 V.
(c) Find the maximum charge that can be stored.
The maximum charge that can be stored in a capacitor is given by the formula:
Q = C * V
Given:
C = 1.40 mF = 1.40 x 10^-3 F
V = 500 V
Substituting the given values:
Q = (1.40 x 10^-3 F) * 500 V
Simplifying the equation:
Q = 0.70 C
Therefore, the maximum charge that can be stored is 0.70 Coulombs.
(d) Calculate the volume of Teflon alone in the capacitor.
The volume of Teflon can be calculated using the formula:
Volume = A * d
Given:
A = 3.16 m²
d = 2.00 µm = 2.00 x 10^-6 m
Substituting the given values:
Volume = 3.16 m² * 2.00 x 10^-6 m
Simplifying the equation:
Volume = 6.32 x 10^-6 m³
Therefore, the volume of Teflon alone in the capacitor is 6.32 x 10^-6 cubic meters.
To solve these problems, we will use the formula for the capacitance of a parallel-plate capacitor:
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 the plates, and d is the separation between the plates.
(a) Calculate the area the parallel plates of such a capacitor must have if they are separated by 2.00 µm of Teflon.
To find the area of the plates, we can rearrange the formula as:
A = (C * d) / ε₀
Given:
C = 1.40 mF = 1.40 x 10^-3 F
d = 2.00 µm = 2.00 x 10^-6 m
ε₀ = 8.85 x 10^-12 F/m
Substituting the given values into the formula:
A = (1.40 x 10^-3 * 2.00 x 10^-6) / (8.85 x 10^-12) = 3.16 m²
Therefore, the area the parallel plates must have is 3.16 m².
(b) What is the maximum voltage that can be applied?
The maximum voltage that can be applied is limited by the breakdown voltage of the dielectric material between the plates, which in this case is Teflon. Let's assume the breakdown voltage of Teflon is V_bd.
The maximum voltage that can be applied without causing breakdown is given by the formula:
V_max = V_bd / √(ε_r)
where V_max is the maximum voltage, V_bd is the breakdown voltage, and ε_r is the relative permittivity (dielectric constant) of the material.
For Teflon, ε_r ≈ 2.1 and V_bd can vary but let's assume it is 400 V for this example.
Substituting the values into the formula:
V_max = 400 / √(2.1) = 275.13 V
Therefore, the maximum voltage that can be applied is approximately 275.13 V.
(c) Find the maximum charge that can be stored.
The maximum charge that can be stored in a capacitor is given by the formula:
Q_max = C * V_max
where Q_max is the maximum charge, C is the capacitance, and V_max is the maximum voltage.
Given:
C = 1.40 x 10^-3 F
V_max = 275.13 V
Substituting the values into the formula:
Q_max = (1.40 x 10^-3 F) * (275.13 V) = 0.385 C
Therefore, the maximum charge that can be stored is 0.385 C.
(d) Calculate the volume of Teflon alone in the capacitor.
To calculate the volume of Teflon alone in the capacitor, we need the area and thickness of Teflon.
Given:
A = 3.16 m² (calculated in part a)
d = 2.00 µm = 2.00 x 10^-6 m
The volume of Teflon can be calculated by multiplying the area and thickness:
Volume = A * d
Volume = (3.16 m²) * (2.00 x 10^-6 m)
Volume = 6.32 x 10^-6 m³
Therefore, the volume of Teflon alone in the capacitor is approximately 6.32 x 10^-6 m³.