mat-sln-asn-conddiel - A-E A = Q enc = b A (c) Substitute...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Solution for Lab Quiz 10 Conductors and Dielectrics Quiz Solution to Lab Quiz Problem 10.1(Compute Bound Charge of Plexiglass) Problem: A piece of plexiglass ( κ = 3 . 2) is placed above an inFnite plane of charge with charge density σ = - 0 . 1 μ C m 2 . Compute the bound charge density on the bottom surface of the dielectric. Be careful of the sign. Select One of the ±ollowing: (a) - 1 . 7 × 10 - 8 C / m 2 (b) 1 . 7 × 10 - 8 C / m 2 (c) - 3 . 4 × 10 - 8 C / m 2 (d-Answer) 3 . 4 × 10 - 8 C / m 2 (e) - 5 . 5 × 10 - 6 C / m 2 - - - - - - - σ b σ t κ = 3.2 Solution (a) Calculate the Fields: Outside the slab, the Feld is that of an inFnite plane. So immediately below the slab 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
the feld is v E 0 = - σ 2 ε 0 ˆ z The feld inside the slab is reduced by a Factor oF κ , v E κ = - σ 2 ε 0 κ ˆ z - - - - - - - - - - + + + + z σ t σ b κ = 3.2 (b) Apply Gauss’ Law: Use a Gaussian Pillbox, a cylinder with end area A as drawn above. The charge enclosed in the cylinder is Q enc = σ b A . Applying Gauss’ Law to the pillbox yields E t A - E b A = E
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: A-E A = Q enc = b A (c) Substitute the Fields: Cancelling A and substi-tuting the feld, p- 2 P-p- 2 P = b b =- 2 + 2 = 2 p-1 + 1 P 2 b = . 1 C m 2 2 p-1 3 . 2 + 1 P = 3 . 4 10-8 C / m 2 2 point(s) : Correct Sign 2 point(s) : Correct Value Total Points for Problem: 4 Points Solution to Lab Quiz Problem 10.2(Dielectric Constant from Field Map) Problem: The fgure to the right shows a dielectric slab in a uniForm electric feld. What is the dielec-tric constant oF the slab? Select One oF the ollowing: (a) (b) 1 (c-Answer) 2 (d) 3 (e) 4 Dielectric E E E Solution Since the feld is thinned by a Factor oF 2 , = 2 . 3 Total Points for Problem: 1 Points 4...
View Full Document

This note was uploaded on 03/29/2010 for the course PHYS 2469 taught by Professor Stewat during the Spring '10 term at University of Arkansas Community College at Batesville.

Page1 / 4

mat-sln-asn-conddiel - A-E A = Q enc = b A (c) Substitute...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online