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

1 answer

(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.