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# Session 2 c palmer r r q 4

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Unformatted text preview: R r Physics 2B - Summer Session 2 - C. Palmer Q ρ= 4 π R3 3 August 12, 2013 10 ˆˆ n=r   Apply Gauss’s Law ∫ QEnclosed E ⋅ dA = ε0   Left hand side ∫ 2 E ⋅ dA = E ( r ) 4π r   Right hand side +Q ˆ E (r ) r r R 3 QEnclosed Q r = 3 ε0 ε0 R Physics 2B - Summer Session 2 - C. Palmer Q ρ= 4 π R3 3 August 12, 2013 11 ˆˆ n=r   Apply Gauss’s Law ∫ QEnclosed E ⋅ dA = ε0   Together 3 +Q ˆ E (r ) r Qr E ( r ) 4π r = 3 ε0 R r R 2 Qr ⇒ E (r ) = 3 4πε 0 R Physics 2B - Summer Session 2 - C. Palmer Q ρ= 4 π R3 3 August 12, 2013 12 ˆˆ n=r   Outside the sphere has an +Q electric field just like a charge. ⎧Q ⎪ 4πε ⎪ 0 E (r ) = ⎨ Q ⎪ ⎪ 4πε 0 ⎩ r ˆ r 3 R r<R 1 ˆ r 2 r r>R ˆ E (r ) r Physics 2B - Summer Session 2 - C. Palmer r R August 12, 2013 13   What will the charge be on the inner wall of the spherical cavity?   What will be the distribution of that charge? +Q Physics 2B - Summer Session 2 - C. Palmer August 12, 2013 14   Total flux is 0 when the surface is in the conductor   Thus the enclosed charge is 0   What will the charge be on the inner wall of the spherical cavity? ∫ ⇒ QEnclosed = 0   –Q   What will be the distribution of that charge?   Not uniform   More negative charge close to the charge   The charge is not centered E ⋅ dA = 0 Physics 2B - Summer Session 2 - C. Palmer - - - +Q - - August 12, 2013 - - - 15   The charge be on the OUTER wall of the conductor will be +Q because it is net neutral.   What will be the distribution of that charge be? - - - +Q - - Physics 2B - Summer Session 2 - C. Palmer August 12, 2013 - - - 16   The outer cavity has no information about the distribution of charge or amount of charge inside the cavity.   This is because the electric field between them is 0.   So the field outside will be the same as if there were simply an excess of charge, +Q. + + - - - +Q + -   the distribution is uniform + - on a conducting sphere. Physics 2B - Summer Session 2 - C. Palmer August 12, 2013 - + - + - + + 17 Physics Concept Components Learning Outcome Evaluate the forces of charged objects on one another. Charge and its analogy Evaluate value and magnitude of electric to mass fields at arbitrary points in space. Coulomb’s Law and its Evaluate the potential differences analogy to gravity between objects (surfaces and points). Electric Force and Electric Fields Calculate flux with surface integrals. Fields Electric Flux Deduce electric fields using Gauss’s Law. Gauss’s Law Discern between electric potential and Electric Potential electric potential energy Capacitance Compute electric field and potential for Material Properties point...
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## This note was uploaded on 09/10/2013 for the course PHYS 2B 2b taught by Professor Hirsch during the Summer '10 term at UCSD.

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