Finding electric field at a point in a capacitor to find capacitance

[hcpark]

Homework Statement

Find the electric field in a capacitor with dielectric constant ε, then calculate the capacitance.

For a parallel plate capacitor, two methods are usually offered.
(M1) Add the electric fields from two infinite charge sheets, one with surface charge density σ, the other with -σ, using principle of superposition, getting E= σ/2ε + σ/2ε = σ/ε.
(M2) Apply Gaussian surface around one of the metal plate to get D=σ, using Gauss's law, as the flux density D inside a metal is zero, then get E=D/ε=σ/ε.

I am confused and need help understanding the following points.
(1) Why is the principle of superposition not applied in the second method(M2), yet we get the same answer? We are still trying to get the electric field at a point, but we get D (and E) from one charge distribution, but not the other. Why is this?
(2) In contrast, for a parallel-wire capacitor, with two long (length l) and thin (radius a) wires carrying line charge density of +λ and -λ, separated by a distance d where a<<d, all the explanations I found use the superposition(M1) to get E=λ/2περ + λ/2πε(d-ρ), where ρ is the distance from one of the wires. But if I apply Gaussian surface to surround one of the wires, I get D=λ/2πρ, E=λ/2περ. Capacitance C will be apparently different from different electric field E. What did I do wrong in this calculation? (Please see my attempt in 3 below.)

Homework Equations

Div D=ρ
V=-∫Edl
C=Q/V

The Attempt at a Solution

For a parallel-wire capacitor, assuming line charge density λ on one wire with the center on z-axis (ρ=0), and -λ on the other wire centered at ρ=d, I applied a Gaussian surface of a pillbox around the wire at ρ=0. I get D=λ/2πρ and E=λ/2περ. Integrating E from d-a to a, V=(λ/2πε)ln(d/a -1), and C=2πε/ln(d/a -1). This is about a factor of 2 different from the "correct" solution from Wikipedia, etc, using E=λ/2περ + λ/2πε(d-ρ) instead.
What is the difference between this problem and a parallel-plate capacitor. Both methods produced same E and same C for the parallel-plate capacitor, but not for the parallel-wire capacitor. Why? Please help.

 

[anathema]

Thank you for your post. I understand your confusion and will try my best to explain the differences between the two methods and why they produce different results for the parallel-wire capacitor.

To start, let's review the two methods for finding the electric field in a capacitor with dielectric constant ε. In method 1 (M1), we use the principle of superposition to add the electric fields from two infinite charge sheets, one with surface charge density σ and the other with -σ. This results in an electric field of E= σ/ε + σ/ε = σ/ε. In method 2 (M2), we apply Gauss's law to a Gaussian surface around one of the metal plates, which gives us D=σ and E=D/ε=σ/ε.

Now, let's look at how these methods are applied to a parallel-wire capacitor. In M1, we still use the principle of superposition, but instead of infinite charge sheets, we have two long, thin wires with line charge density λ. This results in an electric field of E=λ/2περ + λ/2πε(d-ρ), where ρ is the distance from one of the wires. This method takes into account the contributions from both wires, resulting in a different electric field than in the parallel-plate capacitor case.

In M2, we apply Gauss's law to a Gaussian surface around one of the wires. However, unlike in the parallel-plate case, the electric field is not constant over the surface of the Gaussian surface. This is because the wires have a finite length and the electric field varies with distance from the wire. Therefore, the electric field inside the Gaussian surface is not equal to σ/ε, but rather varies depending on the distance from the wire. This is why the electric field and capacitance calculated using M2 are different from the values obtained using M1.

To summarize, the main difference between the two methods is that M1 takes into account the contributions from both wires, while M2 only considers the electric field from one wire at a time. This results in different electric fields and capacitances for the parallel-wire capacitor.

I hope this helps clarify your confusion. Please let me know if you have any further questions.