How to Calculate the Length of Strips for a Parallel-Plate Capacitor
Calculating the Length of Strips for a Parallel-Plate Capacitor
The length of the strips should be 1.87 cm in order to achieve the desired capacitance of 8.00 x 10⁻⁸ F.
The capacitance of a parallel-plate capacitor can be calculated using the formula:
C = ε₀A/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.
In this case, the area of the plates is the product of the width and length of the strips, or A = (4.60 cm)(L), where L is the length of the strips. The distance between the plates is the thickness of the paper, or d = 0.0350 mm.
Since the paper has a dielectric constant of 3.70, we can multiply the permittivity of free space by the dielectric constant to get the effective permittivity of the paper, or ε = ε₀(3.70).
Plugging these values into the formula for capacitance, we get:
C = ε₀(3.70)(4.60 cm)(L)/(0.0350 mm)
Rearranging to solve for L, we get:
L = (C)(0.0350 mm)/[ε₀(3.70)(4.60 cm)]
Plugging in the given values for C, ε₀, and the conversion factors for cm and mm, we get:
L = (8.00 x 10⁻⁸ F)(0.0350 mm)/[(8.85 x 10⁻¹² F/m)(3.70)(4.60 cm)(10⁻² m/cm)(10³ mm/m)]
L = 0.001869 m = 1.87 cm
Therefore, the length of the strips should be 1.87 cm in order to achieve the desired capacitance of 8.00 x 10⁻⁸ F.
Consider a plane parallel-plate capacitor made of two strips of aluminum foil separated by a layer of paraffin-coated paper. Each strip of foil and paper is 4.60 cm wide. The foil is 0.00400 mm thick, and the paper is 0.0350 mm thick and has a dielectric constant of 3.70. What length should the strips be if a capacitance of 8.00 x 10⁻⁸ F is desired? The length of the strips should be 1.87 cm in order to achieve the desired capacitance of 8.00 x 10⁻⁸ F.