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Hw: All Chapter 5 problems and exercises
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Test 1 results Average 75 Median 78 >90>80>70>60>50<50 521 – 79 522 – 76 523 – 73.3 524 – 76.5 525 – 76 526 – 69.7
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Outline Applications of Gauss’s Law - The single Fixed Charge -Field of a sphere of charge -Field of a spherical shell -A Line of Charge Conductors and Insulators The electric field of a conductor The field in the cavity of a conductor; Faraday’s Cage
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Solid conducting sphere with charge Q A E r V r A
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Electric field of a ball of charge Q Electric field outside of a charged sphere is exactly the same as the electric field produced by a point charge, located at the center of the sphere, with charge equal to the total charge on the sphere.
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Electric field of a spherical shell Q The field outside the shell is like that of a point charge, while the field everywhere inside the shell is zero.
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Electric field of a line of charge
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A Charged, Thin Sheet of Insulating Material + + + + + + + + + + +
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Conductors and insulators Charges reside at the surface of the conductor Conductor E=0 + ++ + + + + + + + + + + + +
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What have we learned about conductors? There is no electric field inside a conductor Net charge can only reside on the surface of a conductor Any external electric field lines are perpendicular to the surface (there is no component of electric field that is tangent to the surface). The electric potential within a conductor is constant
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Electric field near a surface of a conductor a
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d + + + + + + a - - - (the total field at any point between the plates) Two parallel conducting plates
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An Apparent Contradiction + + + + + + - - -
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+ + + + + + - - - Near the surface of any conductor in electrostatics
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1) There is a conducting spherical shell, inner radius A and outer radius B. If you put a charge Q on it, find the charge density everywhere. 2) There is a conducting spherical shell, inner radius A and outer radius B. A charge Q is put at the center. If you put a charge Q 2 on the shell, find the charge density everywhere.
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A sphere of radius A has a charge Q uniformly spread throughout its volume. Find the difference in the electric potential, in other words, the voltage difference, between the center and a point 2A from the center.
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There is a conducting spherical shell, inner radius A and outer radius B. A charge Q 1 is put at the center. If you now put charge -2Q 1 on the shell, find the charge density at r=A and r=B.
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since inside the conductor. For any two points and inside the conductor The conductor’s surface is an equipotential.
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Equipotential Surfaces An equipotential surface is a surface on which the electric potential V is the same at every point. Because potential energy does not change as a test charge moves over an equipotential surface, the electric field can do no work on such a charge. So, electric field must be perpendicular to the surface at every point so that the electric force is always perpendicular to the displacement of a charge moving on the surface. Field lines and equipotential surfaces are always mutually perpendicular.
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Method of images: What is a force on the point charge near a conducting plate? - - - - - - - - - - - - - Equipotential surface
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The force acting on the positive charge is exactly the same as it would be with the negative image charge instead of the plate. The point charge feels a force towards the plate with a magnitude:
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Method of images: A point charge near a conducting plane. - - - - - - - - - - - - - Equipotential surface
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Equilibrium in electrostatic field: Earnshaw’s theorem There are NO points of stable equilibrium in any electrostatic field! How to prove it? Gauss’s Law will help! P Imaginary surface surrounding P If the equilibrium is to be a stable one, we require that if we move the charge away from P in any direction, there should be a restoring force directed opposite to the displacement. The electric field at all nearby points must be pointing inward – toward the point P. But that is in violation of Gauss’ law if there is no charge at P.
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Thomson’s atom 1899 If charges cannot be held stably, there cannot be matter made up of static point charges (electrons and protons) governed only by the laws of electrostatics. Such a static configuration would collapse!
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Capacitors Consider two large metal plates which are parallel to each other and separated by a distance small compared with their width. Area A The field between plates is L
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The capacitance is:
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