Physics 212 Lecture 6, Slide 1 Physics 212 Lecture 6 Today's Concept: Electric Potential Defined in terms of Path Integral of Electric Field.

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Physics 212 Lecture 6, Slide 1 Physics 212 Lecture 6 Today's Concept: Electric Potential Defined in terms of Path Integral of Electric Field

Physics 212 Lecture 6 Music Who is the Artist? A)John Prine B)Little Feat C)Taj Mahal D)Ry Cooder E)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 Also, this album is great…. NOTE: Ellnora starts tonight…..

Physics 212 Lecture 6, Slide 3 Your Comments 05 “.” “I'm pretty sure my head just exploded. Just... everywhere.” “help! i need somebody help! not just anybody help! you know i need someone HELLPP" “ ” “That last one was a doozy. Equipotential lines seem like they hold the key to something, but I don't know what yet.” “Do we have to be able to use/do problems with gradients?” “Are we going to differentiate the electric potential in three dimensions in order to get the electric field?” “The calculations with the spherical insulator were hard to follow. And I understand simple ideas, but I hope that the lecture helps me understand more fully.” Electric potential is related to energy – a key aspect of E’nM. We will use electric potential extensively when we talk about circuits. We really only need to know about derivatives (partial derivatives in a few cases). See example at end of class. Not too bad. After discussion today, I understand Gauss's Law!

Physics 212 Lecture 6, Slide 4 BIG IDEA Last time we defined the electric potential energy of charge q in an electric field: 40 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. Note the similarity to the definition of another quantity which is also a PROPERTY OF THE SPACE, the electric field.

Physics 212 Lecture 6, Slide 5 Electric Potential from E field Consider the three points A, B, and C located in a region of constant electric field as shown. 40 What is the sign of  V AC = V C - V A ? (A)  V AC 0 Remember the definition: Choose a path (any will do!) D xxxx

When you integrate 0, you get 0, everywhere. Physics 212 Lecture 6, Slide 6 Checkpoint 2 08 A B D Remember the definition V is constant !! When the electric field is zero in a certain region, the electric potential is surely zero in that region. However it is not certain that the electric potential is zero elsewhere. Suppose the electric field is zero in a certain region of space. Which of the following statements best describes the electric potential in this region of space? A. The electric potential is zero everywhere in this region B. The electric potential is zero at at least one point in this region C. The electric potential is constant everywhere in this region D. There is not enough information to distinguish which of the answers is correct The potential is the integral of E dx. If E is zero then when you integrate you will just get a constant.

Physics 212 Lecture 6, Slide 7 E from V If we can get the potential by integrating the electric field: 40 We should be able to get the electric field by differentiating the potential?? In Cartesian coordinates:

Physics 212 Lecture 6, Slide 8 Checkpoint 1a 08 How do we get E from V?? Look at slopes !!! ““ “The highest electric potential also means the greatest electric field.“ ““ “The slope is the steepest there“ ““ “The field and the potential are inversely related“ The electric potential in a certain region is plotted in the following graph: At which point is the magnitude of the E-field greatest? A.B.C.D.

Physics 212 Lecture 6, Slide 9 Checkpoint 1b 08 A B C D How do we get E from V?? Look at slopes !!! “Because B is along the negative slope on the graph.“ “The slope of the electric potential at point C is positive“ The electric potential in a certain region is plotted in the following graph: At which point is the direction of the E-field along the negative x-axis? A.B.C.D.

Physics 212 Lecture 6, Slide 10Equipotentials Equipotentials are the locus of points having the same potential 40 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

Physics 212 Lecture 6, Slide 11 Checkpoint 3a 08 A B C D “The field lines are dense at points A, B, and C, which indicates that the field is strong. However the lines are very spread out at D, indicating that the field is weakest there..“ The field-line representation of the E-field in a certain region of space is shown below. The dashed lines represent equipotential lines. At which point in space is the E-field weakest? A.B.C.D.

Physics 212 Lecture 6, Slide 12 Checkpoint 3b 08 A B C D A B C D “The equipotential is weaker in the region between c and d than in the region from a to b.“ “The work is directly proportional to the distance between two charges. The distance between C and D is further away from that of A and B.” “The charge would have to cross the same number of equipotential lines regardless, so the change in potential energy is the same and therefore the work is the same.“ The field-line representation of the E-field in a certain region of space is shown below. The dashed lines represent equipotential lines. Compare the work done moving a negative charge from A to B and from C to D. Which move requires more work? A. From A to BB. From C to D C. The sameD. Cannot determine without performing calculation

Physics 212 Lecture 6, Slide 13HINT 08 A B C D ELECTRIC FIELD LINES !! What are these ? EQUIPOTENTIALS !! What is the sign of W AC = work done by E field to move negative charge from A to C ? (A) W AC 0 A and C are on the same equipotential W AC = 0 !! Equipotentials are perpendicular to the E field: No work is done along an equipotential The field-line representation of the E-field in a certain region of space is shown below. The dashed lines represent equipotential lines. What are these ?

Physics 212 Lecture 6, Slide 14 Checkpoint 3b Again? 08 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 The field-line representation of the E-field in a certain region of space is shown below. The dashed lines represent equipotential lines. Compare the work done moving a negative charge from A to B and from C to D. Which move requires more work? A. From A to BB. From C to D C. The sameD. Cannot determine without performing calculation

Physics 212 Lecture 6, Slide 15 Checkpoint 3c 08 A B C D A B C D “because the electric field is constant, moving the charge from A to D would require the most work since it has the furthest displacement.” “The electric field is greater at point D than B.“ “B and D are on the same equipotential line.“ The field-line representation of the E-field in a certain region of space is shown below. The dashed lines represent equipotential lines. Now compare the work done moving a negative charge from A to B and from A to D. Which move requires more work? A. From A to BB. From A to D C. The sameD. Cannot determine without performing calculation

Physics 212 Lecture 6, Slide 16 Calculation for Potential Point charge q at center of concentric conducting spherical shells of radii a 1, a 2, a 3, and a 4. The inner shell is uncharged, but the outer shell carries charge Q. What is V as a function of r? +q metal +Q a1a1 a2a2 a3a3 a4a4 Conceptual Analysis: –Charges q and Q will create an E field throughout space – Strategic Analysis: –Spherical symmetry: Use Gauss’ Law to calculate E everywhere –Integrate E to get V cross-section 40

Physics 212 Lecture 6, Slide 17 Calculation: Quantitative Analysis r > a 4 : What is E(r)? (A) 0 (B) (C) +q metal +Q a1a1 a2a2 a3a3 a4a4 cross-section (D) (E) Why? Gauss’ law: r

Physics 212 Lecture 6, Slide 18 Calculation: Quantitative Analysis a 3 < r < a 4 : What is E(r)? (A) 0 (B) (C) +q metal +Q a1a1 a2a2 a3a3 a4a4 cross-section (D) (E) How is this possible??? -q must be induced at r=a 3 surface -q must be induced at r=a 3 surface r Applying Gauss’ law, what is Q enclosed for red sphere shown? (A) q (B) –q (C) 0 charge at r=a 4 surface = Q+q

Physics 212 Lecture 6, Slide 19 Calculation: Quantitative Analysis Continue on in…. a 2 < r < a 3 : +q metal +Q a1a1 a2a2 a3a3 a4a4 cross-section r a 1 < r < a 2 : r < a 1 : To find V: 1)Choose r 0 such that V(r 0 ) = 0 (usual: r 0 = infinity) 2)Integrate !! r > a 4 : a 3 < r < a 4 : (A) (B) (C)

Physics 212 Lecture 6, Slide 20 Calculation: Quantitative Analysis +q metal +Q a1a1 a2a2 a3a3 a4a4 cross-section r > a 4 : a 3 < r < a 4 : a 2 < r < a 3 : a 1 < r < a 2 :0 < r < a 1 :

Summary V 0 r a1a1 a2a2 a3a3 a4a4