Thesurfaceareaofbothofthecylinders

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Unformatted text preview: surface S 23 rr Integral of E ×dA on surface S Physics 212 Lecture 3, Slide 10 Physics 212 Lecture 3, Slide BB Preflight 2 “Case 1 is bigger because according to Gauss's Law, flux is determined by the charge enclosed.” “The surface area of both of the cylinder's are the same so the flux is the same in both.” “The surface area of the cylinder in case 2 is larger than case 1 so the flux of case 2 is larger. “ 60 Φ1=2Φ2 (A) Φ1=Φ2 (B) Φ1=1/2Φ2 (C) none (D) 50 40 30 20 10 0 25 Physics 212 Lecture 3, Slide 11 Physics 212 Lecture 3, Slide Preflight 2 Definition of Flux: Φ≡ rr ∫ E ⋅ dA surface E constant on barrel of cylinder E perpendicular to barrel surface (E parallel to dA) r Φ = E ∫ dA = EAbarrel barrel Case 1 E1 = λ 2πε 0 r A1 = (2πr ) L 25 Φ1=2Φ2 (A) Φ1 = λL ε0 (B) Φ1=1/2Φ2 (C) none (D) RESULT: GAUSS’ LAW Φ proportional to charge enclosed ! Case 2 E2 = Φ1=Φ2 λ 2πε 0 R A2 = (2πR) L / 2 = πRL Φ2 = λ ( L / 2) ε0 Physics 212 Lecture 3, Slide 12 Physics 212 Lecture 3, Slide Direction Matters: E E E dA For a closed surface, For a closed surface, A points outward dA E dA E E E dA dA E E rr Φ S = ∫ E ×dA > 0 S 26 Physics 212 Lecture 3, Slide 13 Physics 212 Lecture 3, Slide Direction Matters: E E E dA For a closed surface, For a closed surface, A points outward dA E dA E E dA dA E E E rr Φ S = ∫ E ×dA < 0 S 27 Physics 212 Lecture 3, Slide 14 Physics 212 Lecture 3, Slide BB Trapezoid in Constant Field y La b e l fa c e s : 1 : x = 0 2 : z = +a 3 : x = +a 4 : s la nte d rr ˆ E0 E = E0 x 4 1 3 2 x De...
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This document was uploaded on 02/13/2014.

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