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mat-sln-asn-conddiel

# mat-sln-asn-conddiel - κ A-E A = Q enc ε = σ b A ε(c...

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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

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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

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Unformatted text preview: κ A-E A = Q enc ε = σ b A ε (c) Substitute the Fields: Cancelling A and substi-tuting the feld, p-σ 2 ε κ P-p-σ 2 ε P = σ b ε σ b =-σ 2 κ + σ 2 = σ 2 p-1 κ + 1 P 2 σ b = . 1 μ 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 κ E E Solution Since the feld is thinned by a Factor oF 2 , κ = 2 . 3 Total Points for Problem: 1 Points 4...
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mat-sln-asn-conddiel - κ A-E A = Q enc ε = σ b A ε(c...

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