Question

a student decides to construct a capacitor using two conducting plates 30cm on each side. He seprates the plate with a paper dielectric 2.5 that is 8*10 raise to the power -5 mmerters thick. What is the capacitance of the capacitor.

Answer 1

Please type your subject in the School Subject box. Any other words, including obscure abbreviations, are likely to delay responses from a teacher who knows that subject well.

Answer 2

C = epsilon0*K* A/d

where K is the dielectric constant of paper.
d is the plate separation in m
A is the plate area in m^2
epsilono = 8.85*10^-12 Farad/m
You will have to look up K, which will be a dimensionless number greater than 1.

Answer 3

[AI-generated answer — Step-by-Step Bot]

To find the capacitance of the capacitor, we can use the formula:

C = (ε₀ * εᵣ * A) / d,

where:
C is the capacitance,
ε₀ is the vacuum permittivity (8.854 x 10^-12 F/m),
εᵣ is the relative permittivity of the dielectric material,
A is the area of the plates, and
d is the distance between the plates.

Given information:
Side length of each plate (L) = 30 cm = 0.3 m,
Paper dielectric thickness (d) = 8 x 10^-5 m,
Relative permittivity of paper (εᵣ) = 2.5.

First, we need to calculate the area of one plate (A):
A = L^2 = (0.3 m)^2 = 0.09 m^2.

Now, we can substitute the values into the formula:
C = (8.854 x 10^-12 F/m * 2.5 * 0.09 m^2) / (8 x 10^-5 m).

Cancelling units and simplifying:
C = (22.134 x 10^-12 F) / (8 x 10^-5 m).

Dividing the numerator and denominator by 10^-12:
C = 2.767 x 10^-9 F / (8 x 10^-5 m).

Dividing the numerator and denominator by 10^-5:
C = 2.767 x 10^-4 F.

Therefore, the capacitance of the capacitor is 2.767 x 10^-4 Farads.

Answer 4

[AI-generated answer — Explain Bot]

To calculate the capacitance of the capacitor, you need to use the formula for the capacitance of a parallel plate capacitor:

C = (ε₀ * A) / d

Where:
C is the capacitance,
ε₀ (epsilon naught) is the permittivity of free space (approximately 8.854 x 10^-12 F/m),
A is the area of one of the conducting plates, and
d is the distance between the plates.

Let's find the values for the given information:

Plate area, A = (30 cm)^2 = (0.3 m)^2 = 0.09 m^2
Distance between the plates, d = 8 * 10^-5 m

Now, substitute the values into the formula:

C = (8.854 x 10^-12 F/m * 0.09 m^2) / (8 x 10^-5 m)

Simplifying the equation:

C = 8.854 x 10^-12 F * 0.09 m / 8 x 10^-5 m
C = 79.686 x 10^-12 F / 8 x 10^-5 m
C = 995.825 x 10^-12 F

Therefore, the capacitance of the capacitor is approximately 995.825 picofarads (pF).