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W04D1 Electric Potential and Gauss ’ Law Equipotential Lines. Today ’ s Reading Assignment Course Notes: Sections 3.3-3.4, 4.4-4.6. 4.10.5. Exam One Thursday Feb 28 7:30-9:30 pm Room Assignments (See Stellar webpage announcements) Review Tuesday Feb 26 from 9-11 pm in 26-152
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W04D1 Electric Potential and Gauss’ LawEquipotential Lines Today’s Reading Assignment Course Notes: Sections 3.3-3.4, 4.4-4.6. 4.10.5
Exam One Thursday Feb 28 7:30-9:30 pm Room Assignments (See Stellar webpage announcements) Review Tuesday Feb 26 from 9-11 pm in 26-152 PS 3 due Tuesday Tues Feb 26 at 9 pm in boxes outside 32-082 or 26-152 Announcements
Outline Continuous Charge Distributions Review E and V Deriving E from V Using Gauss’s Law to find V from E Equipotential Surfaces
Continuous Charge Distributions Break distribution into infinitesimal charged elements of charge dq. Electric Potential difference between infinity and P due to dq. Superposition Principle: Reference Point:
Group Problem Consider a uniformly charged ring with total charge Q. Find the electric potential difference between infinity and a point P along the symmetric axis a distance z from the center of the ring.
Group Problem: Charged Ring Choose
Electric Potential and Electric Field Set of discrete charges: Continuous charges: If you already know electric field (Gauss’ Law) compute electric potential difference using
Using Gauss’s Law to findElectric Potential from Electric Field If the charge distribution has a lot of symmetry, we use Gauss’s Law to calculate the electric field and then calculate the electric potential V using
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).
Worked Example: Spherical Shells • These two spherical shells have equal but opposite charge. • Find • for the regions • b < r • a < r < b • 0 < r < a • Choose
Electric Potential for Nested Shells From Gauss’s Law r Use Region 1: r > b No field No change in V! 12
Electric Potential for Nested Shells Region 2: a < r < b r Electric field is just a point charge. Electric potential is DIFFERENT – surroundings matter 13
Electric Potential for Nested Shells Region 3: r < a 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
Deriving E from V A = (x,y,z), B=(x+Δx,y,z) Ex = Rate of change in V with y and z held constant
Deriving E from V If we do all coordinates: Gradient (del) operator:
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 From that can you derive E(P)? • Yes, its kQ/a2 (up) • Yes, its kQ/a2 (down) • Yes in theory, but I don’t know how to take a gradient • No, you can’t get E(P) from V(P)
Concept Question Answer: E from V 4. No, you can’t get E(P) from V(P) The electric field is the gradient (spatial derivative) of the potential. Knowing the potential at a single point tells you nothing about its derivative. People commonly make the mistake of trying to do this. Don’t!
Group Problem: E from V • Consider two point like charged objects with charge –Q located at the origin and +Q located at the point (0,a). • Find the electric potential V(x,y)at the point P located at (x,y). • Find the x-and y-components of the electric field at the point P using
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 • larger than that for x < 0 • smaller than that for x < 0 • equal to that for x < 0
Concept Question Answer: E from V Answer: 2. The magnitude of the electric field for x > 0 is smaller than that for x < 0 The slope is smaller for x > 0 than x < 0 Translation: The hill is steeper on the left than on the right.
Concept Question: E from V The above shows potential V(x). Which is true? • Ex > 0 is positive and Ex < 0 is positive • Ex > 0 is positive and Ex < 0 is negative • Ex > 0 is negative and Ex < 0 is negative • Ex > 0 is negative and Ex < 0 is positive
Concept Question Answer: E from V Answer: 2. Ex > 0 is positive and Ex < 0 is negative E is the negative slope of the potential, positive on the right, negative on the left, Translation: “Downhill” is to the left on the left and to the right on the right.
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?
Equipotential Curves: Two Dimensions All points on equipotential curve are at same potential. Each curve represented by V(x,y) = constant
Direction of Electric Field E E is perpendicular to all equipotentials Constant E field Point Charge Electric dipole
Direction of Electric Field E E is perpendicular to all equipotentials Field of 4 charges Equipotentials of 4 charges http://web.mit.edu/viz/EM/visualizations/electrostatics/InteractingCharges/zoo/zoo.htm
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 equipotential