Lect03 - Physics 212 Lecture 3 Today's Concepts: Electric...

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Unformatted text preview: Physics 212 Lecture 3 Today's Concepts: Electric Flux and Field Lines Gauss’ Law Physics 212 Lecture 3, Slide 1 Physics BB Music Who is the Artist? A) B) C) D) E) Professor Longhair Johnny Adams David Egan Dr. John Allen Toussaint classics WHY? A great piano CD Five year anniversary of Katrina Five New Orleans is special…. New Physics 212 Lecture 3, Slide 2 Physics Your Comments “What exactly is epsilon naught?” “I don't understand what epsilon not was so I didn't really don't understand any of the equations.” understand IT’S JUST A JUST CONSTANT CONSTANT q E =k 2r ˆ r 1 k≡ 4πε 0 q E= r 2ˆ 4πε 0 r 1 k = 9 x 109 N m2 / C2 ε0 = 8.85 x 10-12 C2 / N m2 50 “Lets talk flux. It's the thing I'm shakiest on.” “Calculating Electric Field from Arc of Charge!” 40 30 20 10 “Why is gauss' law so important? Why is flux a useful value?” 04 0 Confused Confident Physics 212 Lecture 3, Slide 3 Physics My Comments • You will need to understand integrals in this course!!! • Forces and Fields are Vectors • Always Draw a Picture First… What do the Forces/Fields Look Like? Prelecture E Lecture Worked Example E E Infinite line of charge Finite line of charge (constant λ) (constant dq ˆ E = ∫k 2 r r Homework WORKS FOR ALL !! WORKS E Arc of charge (constant λ) (constant Finite line of charge (non-constant λ) constant Physics 212 Lecture 3, Slide 4 Physics Electric Field Lines Direction & Density of Lines Direction represent represent Direction & Magnitude of E Point Charge: Direction is radial Density α 1/R2 Density 07 Physics 212 Lecture 3, Slide 5 Physics Electric Field Lines Dipole Charge Distribution: Direction & Density Direction Density much more interesting… 10 simulation Physics 212 Lecture 3, Slide 6 Physics Preflight 8 Preflight 3 100 80 60 40 20 0 “The electric field lines are denser around Q1 so it has a greater magnitude..” 12 simulation Physics 212 Lecture 3, Slide 7 Physics Preflight 10 100 80 60 40 20 0 “They must have opposite signs for the field lines to go from one to to the other. “ 13 Physics 212 Lecture 3, Slide 8 Physics Preflight 12 Preflight 3 100 80 60 40 20 0 “The lines are closer together so the The magnitude of the electric field and point B is higher “ 15 Physics 212 Lecture 3, Slide 9 Physics BB Point Charges “Telling the difference between positive and negative charges while looking at field lines. Does field line density from a certain charge give information about the sign of the charge?” -q +2q What charges are inside the red circle? -Q -2Q +Q +Q +2Q +Q A 18 -Q B C D -Q E Physics 212 Lecture 3, Slide 10 Physics Which of the following field line pictures best represents the electric field from two charges that have the same sign but different magnitudes? A 21 B C BB D simulation Physics 212 Lecture 3, Slide 11 Physics Electric Flux “Counts Field Lines” Electric “Im very confused by the general concepts of flux through surface areas. please help” Φ S = ∫ E ⋅ dA S Flux through surface S 23 23 Integral of E ⋅ dA on surface S Physics 212 Lecture 3, Slide 12 Physics BB Preflight 2 radius TAKE s TO BE RADIUS ! “The first case contains twice as The much charge as the second, therefore it has more flux. “ “the areas of the cylinders are in a the ration of 1/2, the charges, 2/1, the flux should be the same .” “Flux is equal to EA and Case 2 Flux has twice the area.. “ has Φ1=2Φ2 (A) Φ1=Φ2 (B) Φ1=1/2Φ2 (C) none (D) 50 40 30 20 10 0 25 Physics 212 Lecture 3, Slide 13 Physics Preflight 2 “The flux is proportional to the Area The that the field is passing through. Although the radius is twice as long in the second case, it's length is half as long. Both cases have the same surface area that the field passes through, so the fluxes are equal. ” Definition of Flux: Φ≡ ∫ E ⋅ dA surface E constant on barrel of cylinder constant E perpendicular to barrel surface (E parallel to dA) Φ1=2Φ2 λ E1 = 2πε 0 s A1 = (2πs ) L 25 Φ1 = λL ε0 E2 = (C) none (D) RESULT: GAUSS’ LAW Φ proportional to charge enclosed ! barrel Case 2 Φ1=1/2Φ2 (B) (A) Φ = E ∫ dA = EAbarrel Case 1 Φ1=Φ2 λ 2πε 0 (2 s ) A2 = (2π (2s)) L / 2 = 2πsL Φ2 = λ ( L / 2) ε0 Physics 212 Lecture 3, Slide 14 Physics Direction Matters: E E E dA For a closed surface, For A points outward dA E E dA E dA dA E E E Φ S = ∫ E ⋅ dA > 0 S 26 Physics 212 Lecture 3, Slide 15 Physics Direction Matters: E E E dA For a closed surface, For A points outward dA E E dA E dA dA E E E Φ S = ∫ E ⋅ dA < 0 S 27 Physics 212 Lecture 3, Slide 16 Physics BB Trapezoid in Constant Field y Label faces: 1: x = 0 2: z = +a 3: x = +a 4: slanted ˆ E0 E = E0 x 4 3 1 2 x Define Φn = Flux through Face n dA E ⋅ dA < 0 z E A A Φ2 < 0 A Φ3 < 0 A Φ4 < 0 B Φ1 = 0 B Φ2 = 0 B Φ3 = 0 B Φ4 = 0 C 32 Φ1 < 0 Φ1 > 0 C Φ2 > 0 C Φ3 > 0 C Φ4 > 0 Physics 212 Lecture 3, Slide 17 Physics BB Trapezoid in Constant Field + Q ˆ E = E0 x y E0 Label faces: 1: x = 0 2: z = +a 3: x = +a 3 +Q 1 2 x Define Φn = Flux through Face n Φ = Flux through Trapezoid z Add a charge +Q at (-a,a/2,a/2) How does Flux change? A A Φ3 increases A Φ increases B Φ1 decreases B Φ3 decreases B Φ decreases C 39 39 Φ1 increases Φ1 remains same C Φ3 remains same C Φ remains same Physics 212 Lecture 3, Slide 18 Physics Gauss Law E E Q E E dA dA E E ∫ closed surface 40 dA E dA ΦS = E dA E E ⋅ dA = Qenclosed ε0 Physics 212 Lecture 3, Slide 19 Physics Preflight 6 (A) ΦE increases (B) ΦE decreases (C) ΦE stays same 100 80 60 40 20 0 43 “It follows from Gauss' Law that the electric flux It through a given, closed surface is proportional to the electric charge that surface encloses. If the charge enclosed withing the surface does not change, neither does the flux of that same surface. It doesn't matter WHERE the charge is located as long as it is enclosed by the surface. . “ Physics 212 Lecture 3, Slide 20 Physics BB Preflight 4 “Although the areas of the sections Although (A) dΦA increases dΦB decreases 60 50 40 (B) dΦA decreases dΦB increases (C) dΦA stays same dΦB stays same are equal in both cases, in the second cases, since the charge is closer to dA, the field is stronger, so the flux will increase on dA, as it will decrease on dB. “ will “Flux is the number of electric field Flux lines passing through the surface elements. In either case, that quantity is the same for both dA and dB.” and 30 20 10 0 41 Physics 212 Lecture 3, Slide 21 Physics Think of it this way: 1 2 The total flux is the same in both cases (just the total number of lines) The flux through the right (left) hemisphere is smaller (bigger) for case 2. 45 Physics 212 Lecture 3, Slide 22 Physics Things to notice about Gauss Law ΦS = ∫ closed surface E ⋅ dA = Qenclosed ε0 If Qenclosed iis the same, the flux has to s be the same, which means that the be which integral must yield the same result for any surface. for 47 Physics 212 Lecture 3, Slide 23 Physics Things to notice about Gauss Law “Gausses law and what concepts are most important that we know .” ∫ E ⋅ dA = Qenclosed ε0 In cases of high symmetry it may be possible to bring E outside outside the integral. In these cases we can solve Gauss Law for E E ∫ dA = EA = Qenclosed ε0 Qenclosed E= Aε 0 So - iif we can figure out Qenclosed and the area of the So f and surface A, then we know E ! then 49 This is the topic of the next lecture… Physics 212 Lecture 3, Slide 24 Physics Final Thoughts…. • Keep plugging away – Prelecture 4 and Preflight 4 due Thursday – Homework 2 due next Tuesday.. More about this next time 50 Physics 212 Lecture 3, Slide 25 Physics ...
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