A sheet of paper 40mm wide and (1.5 x 10^-2) thick is placed between a metal foil of the same width is used to make a 2.0 UF capacitor. If the dielctric constant(relative permitivity of the paper is 2.5 ,what length of paper is required?

width (b) = 40mm = 4x10^-2 m

thickness(d)=1.5x10^-2mm=1.5X10^-5m
K=2.5
c = 2x 10^-6 A/d = 2x 10^-6 F
(εx 2.5x l x4x10^-2 )/1.5X10^-5m =2x 10^-6
l = 33.90 m

Answer please

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To find the length of paper required for a 2.0 uF capacitor, we can use the formula for capacitance:

C = (ε₀ * εᵣ * A) / d

Where:
C is the capacitance in farads (F)
ε₀ is the vacuum permittivity constant (8.85 x 10^-12 F/m)
εᵣ is the relative permittivity or dielectric constant of the material
A is the area of the capacitor plates in square meters (m²)
d is the spacing between the capacitor plates in meters (m)

In this case, the width of the paper is given as 40 mm, which we need to convert to meters:

Width (w) = 40 mm = 40 x 10^-3 m

The thickness of the paper is given as (1.5 x 10^-2) m.

Now, we need to determine the area of the capacitor plates. Since the width of both the paper and the metal foil is the same, the area of the plates will be the product of the width and the length of the paper (l):

A = w * l

To find the length of the paper (l), we rearrange the formula as:

l = A / w

Substituting the known values into the formula, we have:

A = (40 x 10^-3 m) * l

Now, let's substitute the values into the capacitance formula:

2.0 x 10^-6 F = (8.85 x 10^-12 F/m) * (2.5) * (40 x 10^-3 m) * l / (1.5 x 10^-2 m)

Simplifying the equation:

2.0 x 10^-6 F = (8.85 x 10^-12 * 2.5 * 40 x 10^-3 / 1.5 x 10^-2) * l

2.0 x 10^-6 F = 8.85 x 10^-12 * l

To find the value of l, we can rearrange the equation as:

l = (2.0 x 10^-6 F) / (8.85 x 10^-12)

Calculating the length, l:

l = 2.0 x 10^-6 / (8.85 x 10^-12) = 2.26 x 10^5 m

Therefore, the length of paper required for the capacitor is approximately 2.26 x 10^5 meters.

i didnt get it