# Homework Help: Finding electric field at a point in a capacitor to find capacitance

1. Jun 7, 2013

### hcpark

1. The problem statement, all variables and given/known data

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 priciple 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.)

2. Relevant equations
Div D=ρ
V=-∫Edl
C=Q/V

3. 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.

Last edited by a moderator: Jun 8, 2013