Physics212lecture29slide4 electricpotentialgausslaw a

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Unformatted text preview: one by E field Spherical symmetry & Gauss’ law E = 0 inside shell E = 0 inside shell No work done to move q No change in potential energy ! Physics 212 Lecture 29, Slide 4 Electric Potential, Gauss’ Law A thin non-conducting spherical shell of radius a carries a uniformly distributed net surface charge 2Q. A second thin nonconducting shell of radius b carries a uniformly distributed net surface charge 3Q. The two shells are concentric. Calculate the potential difference ∆ V = V(a) – V(b) between the inner and outer shells. 37% (A) 1 2Q 3Q − 4πε 0 a b (B) 1 3Q 2Q − 4πε 0 b a (C) (D) 0 (E) ALWAYS START FROM DEFINITION OF POTENTIAL a ∆ V = − ∫ E ⋅ dr 1 2Q 2Q − 4πε 0 a b 1 2Q 2Q − 4πε 0 b a Spherical symmetry & Gauss’ law determines E 1 2Q a < r < b: E = 4πε0 r 2 b 2Q ∆V = − 4πε0 a dr ∫2 br 2Q 1 1 ∆V = − 4πε0 a b Physics 212 Lecture 29, Slide 5 Electric Potential, Gauss’ Law A thin non-conducting spherical shell of radius a carries a uniformly distributed net surface charge 2Q. A second thin nonconducting shell of radius b carries a uniformly distributed net surface charge 3Q. The two shells are concentric. Calculate the potential difference ∆ V = V(a) – V(0) between the inner shell and the origin. 45% (A) 1 2Q 3Q − 4πε 0 a b (B) 1 3Q 2Q − 4πε 0 b a (C) 0 (E) Spherical symmetry & Gauss’ law determines E r < a: (D) E=0 1 2Q 2Q − 4πε 0 a b 1 2Q 2Q − 4πε 0 b a a ∆V = − ∫ E ⋅ dr = 0 0 Physics 212 Lecture 29, Slide 6 Gauss’ Law, Conductors A solid conducting sphere of radius a is centered on the origin, a...
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This note was uploaded on 02/18/2013 for the course PHYS 212 taught by Professor Kim during the Fall '08 term at University of Illinois, Urbana Champaign.

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