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# Lect06 - Physics 212 Lecture 6 Today's Concept Electric...

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Unformatted text preview: Physics 212 Lecture 6 Today's Concept: Electric Potential Defined in terms of Path Integral of Electric Field Physics 212 Lecture 6, Slide 1 Physics Music Who is the Artist? A) B) C) D) E) BB BB John Prine Little Feat Taj Mahal Ry Cooder Los Lobos Why? Last time did Buena Vista Social Club (Cuba) Ry Cooder was the guy who brought them to our attention in this country Ry country Also, this album is great…. Physics 212 Lecture 6, Slide 2 Physics Your Comments “I don't get questions 9 and 11 on this preflight. It would be great if we could go eat over these in class. over We defInitely will. They had the lowest We scores.. Good to see you recognize that scores.. “Is the homework intentionally harder than our other examples/problems? I hope it's not just to blems? scare us into dropping or something.” scare “This seems confusing in general. Also, can we try to go over problems that are similar to blems what's on the homework? I feel like the homework involves a lot more stuff that what's mentioned in class or in the book.” mentioned “Just do some hard, concrete problems. Please.” Homework is meant to be challenging. A necessary part of the learning process BUT I WILL BE DOING CALCULATIONS IN CLASS! 50 40 30 20 10 0 05 Confused Confident Physics 212 Lecture 6, Slide 3 Physics BIG IDEA • Last time we defined the electric potential energy of charge q in an electric field: b b a a ∆U a →b = − ∫ F ⋅ dl = − ∫ qE ⋅ dl •The only mention of the particle was through its charge q. • We can obtain a new quantity, the electric potential, which is a PROPERTY OF THE SPACE, as the potential energy per unit charge. ∆U a →b ≡ = − ∫ E ⋅ dl q a b ∆Va →b •Note the similarity to the definition of another quantity which is also a PROPERTY OF THE SPACE, the electric field. F E≡ q 40 Physics 212 Lecture 6, Slide 4 Physics Electric Potential from E field • Consider the three points A, B, and C located in a region of constant electric field as shown. D ∆x BB BB • What is the sign of ∆VAC = VC - VA ? (A) ∆VAC < 0 (B) ∆VAC = 0 (C) ∆VAC > 0 C • Remember the definition: ∆VA→C = − ∫ E ⋅ dl A • Choose a path (any will do!) D C A D ∆VA→C = − ∫ E ⋅ dl − ∫ E ⋅ dl 40 C ∆VA→C = 0 − ∫ E ⋅ dl = − E∆x < 0 D Physics 212 Lecture 6, Slide 5 Physics Preflight 6 A B C D BB • Remember the definition 50 40 B ∆VA→ B = − ∫ E ⋅ dl A 30 20 10 0 E =0 08 ∆VA→ B = 0 A B C D V is constant !! Physics 212 Lecture 6, Slide 6 Physics E from V • If we can get the potential by integrating the electric field: b ∆Va →b = − ∫ E ⋅ dl a • We should be able to get the electric field by differentiating the potential?? E = −∇V • In Cartesian coordinates: Ex = − ∂V dx Ey = − ∂V dy ∂V Ez = − dz 40 Physics 212 Lecture 6, Slide 7 Physics Preflight 2 A B C D BB 60 50 40 30 20 10 0 A B C D “The higher the electric potential on a graph at a greater distance, the greater the E-Field. “ “E=-dV/dx. Since the slope is greatest at B, the magnitude of the E-field is greatest at B. “ “E is the derivative of V, and the slope is greatest at point C. “ • How do we get E from V?? E = −∇V 08 Ex = − ∂V dx Look at slopes !!! Physics 212 Lecture 6, Slide 8 Physics Preflight 4 A B C D BB 50 40 30 20 10 0 A B C D “At B, the slope is decreasing (-) so the direction of the Efield is negative “ “E is negative when the slope of V is positive (E=-dV/dx). Therefore E is directed along the x-axis at point C. “ • How do we get E from V?? E = −∇V 08 Ex = − ∂V dx Look at slopes !!! Physics 212 Lecture 6, Slide 9 Physics Equipotentials • Equipotentials are the locus of points having the same potential Equipotentials produced by a point charge Equipotentials are ALWAYS perpendicular to the electric field lines The SPACING of the equipotentials indicates The STRENGTH of the electric field 40 Physics 212 Lecture 6, Slide 10 Physics Preflight 7 A B C D 100 80 60 40 20 0 A B C D “The electric field lines are the least dense at D “ 08 Physics 212 Lecture 6, Slide 11 Physics Preflight 9 A B C D BB 40 30 20 10 0 A B C D A B C D “A to B requires more work because you have to cross field lines instead of just moving along one like it would be from C to D. “ “C to D because it is a farther distance than from A to B. ” “Since A and C share an equipotential line, as do B and D, the work moving from A to C and from B 08 to D is the same. “ Physics 212 Lecture 6, Slide 12 Physics HINT What are these ? A B C D BB ELECTRIC FIELD LINES !! What are these ? EQUIPOTENTIALS !! • What is the sign of WAC = work done by E field to move negative charge from A to C ? (A) WAC < 0 (B) WAC = 0 (C) WAC > 0 A and C are on the same equipotential 08 Equipotentials are perpendicular to the E field: No work is done along an equipotential WAC = 0 !! Physics 212 Lecture 6, Slide 13 Physics Preflight 9 Again? A B C D BB 40 30 20 10 0 A B C D A B C D • A and C are on the same equipotential • B and D are on the same equipotential • Therefore the potential difference between A and B is the SAME as the potential between C and D 08 Physics 212 Lecture 6, Slide 14 Physics Preflight 11 A B C D BB 50 40 30 20 10 0 A B C D A B C D “The distance that A is moving is greater when it is going to D, so it requires more work “ “Again the change in potential is the same, because B and D are on the same equipotential line... so work done is the same. “ 08 Physics 212 Lecture 6, Slide 15 Physics Calculation for Potential cross-section a4 a3 +Q a2 a1 +q metal Point charge q at center of concentric conducting spherical shells of radii a1, a2, a3, and a4. The inner shell is uncharged, but the outer shell carries charge Q. What is V as a function of r? metal • Conceptual Analysis: – – Charges q and Q will create an E field throughout space r V (r ) = − ∫ E id ℓ r0 • Strategic Analysis: – – 40 Spherical symmetry: Use Gauss’ Law to calculate E everywhere Integrate E to get V Physics 212 Lecture 6, Slide 16 Physics Calculation: Quantitative Analysis cross-section a4 a3 r > a4 : What is E(r)? +Q a2 a1 (A) 0 (A) (B) 1Q 4πε r 2 1 Q+q 2πε r (C) +q +q 1 Q+q (D) 4πε r metal r 2 0 BB 0 0 1 Q−q (E) 4πε r 2 0 metal Why? Gauss’ law: ∫ E i dA = Q enclosed ε 0 E 4π r = 2 Q+q ε 0 1 Q+q E= 4πε r 2 0 Physics 212 Lecture 6, Slide 17 Physics Calculation: Quantitative Analysis cross-section a4 a3 a3 < r < a4 : What is E(r)? +Q a2 a1 (A) 0 (A) +q metal 1 q (B) (B) 4πε r 2 0 r 1 −q (D) 4πε r 2 0 1q (C) 2πε r BB 0 1 Q−q (E) 4πε r 2 0 metal Applying Gauss’ law, what is Qenclosed for red sphere shown? law, (B) –q (A) q (A) (C) 0 How is this possible??? -q must be induced at r=a3 surface σ= 3 −q 4π a 2 3 charge at r=a4 surface = Q+q σ= 4 Q+q 4π a 2 4 Physics 212 Lecture 6, Slide 18 Physics Calculation: Quantitative Analysis cross-section a4 a3 +Q Continue on in…. a2 < r < a3 : a2 a1 E= 1 q 4πε r 2 0 +q metal metal r a1 < r < a2 : BB E=0 r < a1 : E= 1 q 4πε r 2 0 To find V: 1) Choose r0 such that V(r0) = 0 2) Integrate !! r > a4 : V= (usual: r0 = infinity) 1 Q+q 4πε r 0 a3 < r < a4 : (A) V =0 (B) V= 1 Q+q 4πε a 1 Q+q V= 4πε a 0 (C) 4 4 0 = ∆V (∞ → a ) + 0 3 Physics 212 Lecture 6, Slide 19 Physics Calculation: Quantitative Analysis cross-section a4 a3 V= r > a4 : +Q 1 Q+q 4πε r 0 a2 a1 a3 < r < a4 : V = +q 1 Q+q 4πε a 0 4 metal metal a2 < r < a3 : V (r ) = ∆V (∞ → a ) + 0 + ∆V (a → r ) 4 V (r ) = Q+q q 1 1 +0+ r −a 4πε a 4πε 0 a1 < r < a2 : V (r ) = 3 4 0 V (r ) = 3 1 Q+q q q a +r −a 4πε 0 4 3 1 Q+q q q a +a −a 4πε 0 4 2 3 1 Q+q q q q q 0 < r < a1 : V (r ) = +−+− 4πε a a a r a 0 4 2 3 1 Physics 212 Lecture 6, Slide 20 Physics ...
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