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 60 50 40 30 20 10 0 Confused Avg = 3.0 Confident Physics 212 Lecture 6, Slide 1 Music Who is the Artist? A) B) C) D) E) Eric Clapton Bill Frisell Bill Frisell Sonny Landreth Pat Metheny Pat Metheny Jeff Beck Why?? I think he is my favorite jazz guitarist He is playing Saturday afternoon at Krannert! He Krannert PS Pat Metheny and Sonny Landreth have Metheny and played at past Krannert guitar festivals Krannert Physics 212 Lecture 6, Slide 2 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 3 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 • Remember the definition C ∆VA→C = − ∫ E ⋅ dl A C • Choose a path (any will do!) D D D C A ∆VA→C = 0 − ∫ E ⋅ dl = − E∆x < 0 ∆VA→C = − ∫ E ⋅ dl − ∫ E ⋅ dl 40 Physics 212 Lecture 6, Slide 4 Preflight 6 A B C D BB 60 50 40 30 20 10 0 A B C D • Remember the definition B ∆VA→ B = − ∫ E ⋅ dl A 08 ∆V A→ B = 0 V is constant !! Physics 212 Lecture 6, Slide 5 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 6 Preflight 2 A B C D BB 60 50 40 30 20 10 0 A B C D “The point with highest electric potential will also have the greatest E-FIELD. “ FIELD. “The E-Field is greatest at pt. B because at that pt the slope is the steepest. “ teepest. “Since at point C the electrical potential energy is lowest, the kinetic energy will be Since highest, and therefore the magnitude of the E-Field will be greatest. “ highest, Field • How do we get E from V?? E = −∇V 08 Ex = − ∂V dx Look at slopes !!! Physics 212 Lecture 6, Slide 7 Preflight 4 A B C D BB 50 40 30 20 10 0 A B C D “The direction of the E-field is along the x-axis at point B because the slope of the V graph at point B is negative. “ “The direction of the E-field is along the negative x-axis at point C because E= -dV/dr so the derivative must be positive for the electric field to point in the negative x direction and this occurs 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 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 9 Preflight 7 A B C D 100 80 60 40 20 0 A B C D “The E-field lines at D are the least dense. “ 08 Physics 212 Lecture 6, Slide 10 10 Preflight 9 A B C D BB 40 30 20 10 0 A B C D A B C D “The E-field lines are more dense from A to B than C to D, so its going to be harder to move a negative charged between the A and B. “ “C to D is farther away so more work will be needed to move it.” “A,C and B,D are on the same equipotential lines, so it will require the same amount of work. “ 08 Physics 212 Lecture 6, Slide 11 11 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 12 12 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 13 13 Preflight 11 A B C D BB 60 50 40 30 20 10 0 A B C D A B C D “Moving from A-D is across a greater distance, 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 14 14 cross-section a4 a3 Homework Problem +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 15 15 Homework Problem: 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 idA = 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 16 16 Homework Problem: 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 (C) 1q 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 surface Q+q σ= 4 Q+q 4π a 2 4 Physics 212 Lecture 6, Slide 17 17 Homework Problem: 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 18 18 Homework Problem: 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 19 19 ...
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This note was uploaded on 09/22/2011 for the course PHYSICS 212 taught by Professor Selig during the Fall '10 term at University of Illinois, Urbana Champaign.

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