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# week3 - Lecture slides are posted each Thursday for that...

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Lecture slides are posted each Thursday for that week. Thursday, September 9, 2010

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What is an electric flux through spheres of radii r1 and r2= 2r1 surrounding a charge of q=-3 e? 2 -3e A : Φ 1 = Φ 2 = 0 B : Φ 1 = Φ 2 = 0 C : Φ 1 = Φ 2 / 2 D : Φ 1 = 2 Φ 2 r1 r2 E : Φ 1 = Φ 2 / 4 Thursday, September 9, 2010
What is a electric flux through spheres of radii r1 and r2= 2r1 surrounding a charge of q=-3 e? 3 -3e A : Φ 1 = Φ 2 = 0 B : Φ 1 = Φ 2 = 0 C : Φ 1 = Φ 2 / 2 D : Φ 1 = 2 Φ 2 r1 r2 equal: A E : Φ 1 = Φ 2 / 4 Thursday, September 9, 2010

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What is the flux through the spherical surface encompassing two charges q1= 3q, q2=-2q iClicker 4 B A C C D q 0 5 q 0 q 0 3 q 0 0 3q -2q Thursday, September 9, 2010
What is the flux through the spherical surface encompassing two charges q1= 3q, q2=-2q iClicker 5 B A C C D q 0 5 q 0 q 0 3 q 0 0 3q -2q q tot = 3 q 2 q = q Thursday, September 9, 2010

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The total electric flux through two closed surfaces is A: equal B: not equal iClicker 6 Thursday, September 9, 2010
iClicker The total flux of electric field through the closed surface is A: > 0 B: =0 C: <0 7 +3q +3q +3q +3q +3q -q Thursday, September 9, 2010

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August 12, 2010 University Physics, Chapter 22 64 Gauss’s Law for Various Charge We have applied Gauss’s Law to a point charge and showed that we get Coulomb’s Law Now let’s look at more complicated distributions of charge and calculate the resulting electric field We will use a “charge density” to describe the distribution of charge This charge density will be different depending on the geometry Thursday, September 9, 2010
August 12, 2010 University Physics, Chapter 22 65 Cylindrical Symmetry (1) Let’s calculate the electric field from a conducting wire with charge per unit length λ using Gauss’s Law We start by assuming a Gaussian surface in the form of a right cylinder with radius r and length L placed around the wire such that the wire is along the axis of the cylinder E · dA = q 0 Thursday, September 9, 2010

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August 12, 2010 University Physics, Chapter 22 66 Cylindrical Symmetry (2) From symmetry we can see that the electric field will extend radially from the wire How? If we rotate the wire along its axis, the electric field must look the same Cylindrical symmetry If we imagine a very long wire, the electric field cannot be different anywhere along the length of the wire Translational symmetry Thus our assumption of a right cylinder as a Gaussian surface is perfectly suited for the calculation of the electric field using Gauss’ Law E Thursday, September 9, 2010
August 12, 2010 University Physics, Chapter 22 67 Cylindrical Symmetry (3) The electric flux through the ends of the cylinder is zero because the electric field is always parallel to the ends The electric field is always perpendicular to the wall of the cylinder so … and now solve for the electric field Thursday, September 9, 2010

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August 12, 2010 University Physics, Chapter 22 68 Planar Symmetry (1) Assume that we have a thin, infinite non-conducting sheet of positive charge
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week3 - Lecture slides are posted each Thursday for that...

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