A parallel plate air capacitor has a capacitance of 4.70 µF: Determine the new capacitance if: a.) The distance between the plates is doubled (everything else stays the same). b.) The area of the plates is duplicated (everything else remains the same except the 4.70 µF level). c.) A dielectric with a relative permittivity of 20 is inserted between the plates (everything else remains the same except the 4.70 µF level). d.) A dielectric with a relative permittivity of 4 is inserted, the area is reduced to 1/3, and the distance to 1/4 of its original dimensions.

Question

A parallel plate air capacitor has a capacitance of 4.70 µF: Determine the new capacitance if: a.) The distance between the plates is doubled (everything else stays the same). b.) The area of the plates is duplicated (everything else remains the same except the 4.70 µF level). c.) A dielectric with a relative permittivity of 20 is inserted between the plates (everything else remains the same except the 4.70 µF level). d.) A dielectric with a relative permittivity of 4 is inserted, the area is reduced to 1/3, and the distance to 1/4 of its original dimensions.

Answer 1

To determine the new capacitance in the given scenarios, we can use the basic 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
d is the distance between the plates

Let's calculate the new capacitance in each scenario:

a) If the distance between the plates is doubled (doubled to 2d), while everything else stays the same, the formula becomes:

C' = ε₀ * (A / 2d)

Since we want to determine the new capacitance, C', we can substitute the original capacitance value (C = 4.70 µF = 4.70 x 10^-6 F) into the equation:

4.70 x 10^-6 F = ε₀ * (A / 2d)

Now, we can rearrange the equation to solve for C':

C' = (4.70 x 10^-6 F) * (2d / A)

b) If the area of the plates is duplicated (doubled to 2A), while everything else remains the same except for the 4.70 µF level, the formula becomes:

C' = ε₀ * (2A / d)

Substituting the original capacitance value (C = 4.70 µF = 4.70 x 10^-6 F), we get:

4.70 x 10^-6 F = ε₀ * (2A / d)

Rearranging the equation to solve for C':

C' = (4.70 x 10^-6 F) * (d / 2A)

c) If a dielectric with a relative permittivity of 20 is inserted between the plates, while everything else remains the same except for the 4.70 µF level, the formula becomes:

C' = (ε₀ * εᵣ) * (A / d)

where εᵣ is the relative permittivity of the dielectric.

Substituting the original capacitance value (C = 4.70 µF = 4.70 x 10^-6 F) and the relative permittivity (εᵣ = 20), we have:

4.70 x 10^-6 F = (ε₀ * 20) * (A / d)

To solve for C', we can rearrange the equation:

C' = (4.70 x 10^-6 F) * (d / (ε₀ * 20A))

d) If a dielectric with a relative permittivity of 4 is inserted, the area is reduced to 1/3, and the distance is reduced to 1/4 of its original dimensions, the formula becomes:

C' = (ε₀ * 4) * ((A / 3) / (d / 4))

Substituting the original capacitance value (C = 4.70 µF = 4.70 x 10^-6 F), the relative permittivity (εᵣ = 4), and the scaled dimensions:

4.70 x 10^-6 F = (ε₀ * 4) * ((A / 3) / (d / 4))

Rearranging the equation to solve for C':

C' = (4.70 x 10^-6 F) * (4d / (3A))

So, using these formulas, you can calculate the new capacitance in each scenario.