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303L-HW11

303L-HW11 - erro(jmf2547 HW11 berk(60290 This print-out...

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fierro (jmf2547) – HW11 – berk – (60290) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Consider the figure + Q #1 + + + + + + + + + + + - Q #2 - - - - - - - - - - - A B C D x y Of the following elements, identify all that correspond to an equipotential line or surface. 1. both AB and CD 2. line AB only correct 3. line CD only 4. neither AB nor CD Explanation: Consider the electric field + Q #1 + + + + + + + + + + + - Q #2 - - - - - - - - - - - A B C D x y An equipotential line or surface ( AB ) is normal to the electric field lines. 002 (part 2 of 2) 10.0 points Consider the figure - A - q + B + q C D Of the following elements, identify all that correspond to an equipotential line or surface. 1. neither AB nor CD 2. line CD only correct 3. line AB only 4. both AB and CD Explanation: Consider the electric field: - A + B C D An equipotential line or surface ( CD ) is normal to the electric field lines. 003 10.0 points A wire that has a uniform linear charge den- sity of 1 . 3 μ C / m is bent into the shape as shown below, with radius 4 m. 8 m 8 m p 4 m The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Find the electrical potential at point p . Correct answer: 62377 . 8 V. Explanation:
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fierro (jmf2547) – HW11 – berk – (60290) 2 Let : λ = 1 . 3 μ C / m = 1 . 3 × 10 - 6 C / m , R = 4 m , 2 R = 8 m , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . 2 R 2 R p R Let p be the origin. Consider the potential due to the line of charge to the right of p . V right = integraldisplay dV = k e integraldisplay d q r = k e integraldisplay 3 R R λ d x x = k e λ ln x vextendsingle vextendsingle vextendsingle vextendsingle 3 R R = k e λ ln 3 . By symmetry, the contribution from the line of charge to the left of p is the same. The contribution from the semicircle is V semi = k e integraldisplay π 0 λ R d θ R = k e λ integraldisplay π 0 d θ = k e λ θ vextendsingle vextendsingle vextendsingle vextendsingle π 0 = k e λ π .
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