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Unformatted text preview: Physics 217: Problem Set 5 Solutions Yao Liu, Panos, and CJF March 2010 1 Purcell 4.05 Ohm’s law gives J 1 = σ 1 E 1 J 2 = σ 2 E 2 The current density along the wire is the same, as long as the cross section of the wire remains constant. As a matter of fact, the integral of J over any cross section should give the current I , which is constant along the wire. Since the current density is also constant across the cross section, I A = J = σ 1 E 1 = σ 2 E 2 so, Δ E = E 2 E 1 = I Aσ 2 I Aσ 1 We know the jump in E field is related to surface charge density by Δ E = 4 πσ or Δ E = σ (Don’t confuse the two quantities that share the same Greek letter σ !) so the total charge accumulated at the junction is Q = σA = Δ E 4 π A = I 4 π 1 σ 2 1 σ 1 or Q = I 1 σ 2 1 σ 1 If σ 2 < σ 1 , Q is positive, which makes sense because the second half of the wire is less “conducting” than the first half, so positive charges accumulate. In an actual wire, of course, it is the negative electrons that move, which move from a less conducting material to a more conducting material, so there is a depletion of electrons at the junction. 2 Purcell 4.18 R i and R are in series, so the total resistance is R i + R . The current is thus I = V R i + R so the power delivered to R is P = I 2 R = V 2 R ( R i + R ) 2 as a function of R . To find the maximum, we differentiate dP dR = V 2 ( R i + R ) 2 2 R ( R i + R ) ( R i + R ) 4 = 0 ⇐⇒ R = R i 1 3 Griffiths 7.1 By Ohm’s law, in the region between the two spheres, J = σ E and if we take the gradient of both sides...
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This note was uploaded on 06/10/2011 for the course PHYS 2217 taught by Professor Leclair, a during the Spring '06 term at Cornell University (Engineering School).
 Spring '06
 LECLAIR, A
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