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MIT 8 02T - Electric Potential and Gauss’ Law Equipotential Lines

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1 1 W04D1 Electric Potential and Gaussʼ Law Equipotential Lines Todayʼs Reading Assignment Course Notes: Sections 3.3-3.4, 4.4-4.6. 4.10.5 Announcements Exam One Thursday Feb 27 7:30-9:30 pm Room Assignments (See Stellar and 8.02t webpage announcements) Review Sessions: (1) Tuesday Feb 25 from 7:30-9:00 pm in 26-152 (2) Tuesday Feb 25 from 9:00-10:30 pm in 26-152 PS 3 due Tuesday Tues Feb 25 at 9 pm in boxes outside 32-082 or 26-152 2 3 Outline Review E and V Deriving E from V Using Gaussʼs Law to find V from E Equipotential Surfaces2 4 Electric Potential and Electric Field Set of discrete charges: Continuous charges: If you already know electric field (Gaussʼ Law) compute electric potential difference using V (!r) ! V (") = ked#q!r -!#rsource$ VB! VA= !!EAB"# d!s V (!r) ! V (") = keqi!r -!rii#5 E Field and Potential: Effects !F = q!EIf you put a charged particle, (charge q), in a field: Wext= !U = q!VTo move a charged particle, (charge q), in a field and the particle does not change its kinetic energy then: 6 Concept Question: Two Point Charges The work done in moving a positively charged object that starts from rest at infinity and ends at rest at the point P midway between two charges of magnitude +Q and –Q 1. is positive. 2. is negative. 3. is zero. 4. can not be determined – not enough info is given.3 7 E Field and Potential: Creating A point charge q creates a field and potential around it: Use superposition for systems of charges V (!r) ! V (") = ked#q!r -!#rsourc e$ !E = keqr2ˆr V (r) ! V (") = keqr8 Using Gaussʼs Law to find Electric Potential from Electric Field If the charge distribution has a lot of symmetry, we use Gaussʼs Law to calculate the electric field and then calculate the electric potential V using BBAAVV d−=−⋅∫Es9 Group Problem: Coaxial Cylinders A very long thin uniformly charged cylindrical shell (length h and radius a) carrying a positive charge +Q is surrounded by a thin uniformly charged cylindrical shell (length h and radius a ) with negative charge -Q , as shown in the figure. You may ignore edge effects. Find V(b) – V(a).4 10 Worked Example: Spherical Shells These two spherical shells have equal but opposite charge. Find for the regions (i) b < r (ii) a < r < b (iii) 0 < r < a Choose V (!) = 0 V (r) ! V (")11 Electric Potential for Nested Shells !E =Q4!"0r2ˆr, a < r < b0, elsewhere#$%&%%From Gauss’s Law V (r ) ! V (")= 0!= ! 0"r#dr = 0 VB! VA= !!EAB"# d!sUse Region 1: r > b r No field  No change in V! 12 Electric Potential for Nested Shells Region 2: a < r < b V (r) ! V (r = b)= 0! "# $#= ! drQ4"#0r2br$ =Q4!"0rr = br =14!"0Q1r#1b$%&'()r Electric field is just a point charge. Electric potential is DIFFERENT – surroundings matter5 13 Electric Potential for Nested Shells Region 3: r < a V (r) ! V (a)= kQ1a!1b"#$%&'!= ! dr 0ar(= 0 V (r) = V (a) =Q4!"01a#1b$%&'()r Again, potential is CONSTANT since E = 0, but the potential is NOT ZERO for r < a. 14 Group Problem: Charge Slab Infinite slab of thickness 2d, centered at x = 0 with uniform charge density . Find ! V (xA) ! V (0) ; xA> d15 Deriving E from V !!s = !xˆiA = (x,y,z), B=(x+Δx,y,z) Ex! "#V#x$ "%V%xEx = Rate of change in V with y and z held constant !V = "!EAB#$ d!s !V = "!E( x, y,z )( x+!x, y,z )#$ d!s % "!E $ !!s = "!E $(!xˆi) = " Ex!x6 16 Gradient (del) operator: !! "##xˆi +##yˆj +##zˆkIf we do all coordinates: !E = !!"VDeriving E from V !E = !"V"xˆi +"V"yˆj +"V"zˆk#$%&'( = !""xˆi +""yˆj +""zˆk#$%&'(V17 Concept Question: E from V Consider the point-like charged objects arranged in the figure below. The electric potential difference between the point P and infinity and is V ( P) = ! kQ aFrom that can you derive E(P)? 1. Yes, its kQ/a2 (up) 2. Yes, its kQ/a2 (down) 3. Yes in theory, but I donʼt know how to take a gradient 4. No, you canʼt get E(P) from V(P) 18 Group Problem: E from V Consider two point like charged objects with charge –Q located at the origin and +Q located at the point (a,0). (a) Find the electric potential V(x,y)at the point P located at (x,y). (b) Find the x-and y-components of the electric field at the point P using Ex(x, y) = !"V"x, Ey(x, y) = !"V" y7 19 Concept Question: E from V The graph above shows a potential V as a function of x. The magnitude of the electric field for x > 0 is 1. larger than that for x < 0 2. smaller than that for x < 0 3. equal to that for x < 0 20 Concept Question: E from V The above shows potential V(x). Which is true? 1. Ex > 0 is positive and Ex < 0 is positive 2. Ex > 0 is positive and Ex < 0 is negative 3. Ex > 0 is negative and Ex < 0 is negative 4. Ex > 0 is negative and Ex < 0 is positive 21 Group Problem: E from V A potential V(x,y,z) is plotted above. It does not depend on x or y. What is the electric field everywhere? Are there charges anywhere? What sign? -5 0 50510Potential (V)Z Position (mm)8 22 Equipotentials 23 Topographic Maps 24 Equipotential Curves: Two Dimensions All points on equipotential curve are at same potential. Each curve represented by V(x,y) = constant9 25 Direction of Electric Field E E is perpendicular to all equipotentials Constant E field Point Charge Electric dipole 26 Direction of Electric Field E http://web.mit.edu/viz/EM/visualizations/electrostatics/InteractingCharges/zoo/zoo.htm E is perpendicular to all equipotentials Field of 4 charges Equipotentials of 4 charges 27 Properties of Equipotentials E field lines point from high to low potential E field lines perpendicular to equipotentials E field has no component along equipotential The electrostatic force does zero work to move a charged particle along equipotential10 28 Group Problem: Equipotential Game Open applet, play with, answer questions on worksheet and hand in answers


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MIT 8 02T - Electric Potential and Gauss’ Law Equipotential Lines

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