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Unformatted text preview: zhou (ejz224) – Homework 3 – shih – (57480) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Four charges are at the corners of a square centered at the origin as follows: q at ( − a, + a ); 2 q at (+ a, + a ); − 3 q at (+ a, − a ); and 6 q at ( − a, − a ). A fifth charge + q with mass m is placed at the origin and released from rest. Find the speed when it is a great distance from the origin. 1. bardbl vectorv bardbl = q radicalBigg 3 √ 6 k ma 2. bardbl vectorv bardbl = q radicalBigg 6 √ 2 k ma correct 3. bardbl vectorv bardbl = q radicalBigg 3 √ 3 k ma 4. bardbl vectorv bardbl = q radicalBigg 6 √ 3 k ma 5. bardbl vectorv bardbl = q radicalBigg 6 √ 6 k ma 6. bardbl vectorv bardbl = q radicalBigg 3 √ 2 k ma Explanation: x a +6 q a − 3 q +2 q + q + q, m √ 2 a The initial energy of the charge is E i = K i + U i = U i = q parenleftbigg k q √ 2 a + 2 k q √ 2 a + ( − 3 q ) k √ 2 a + 6 k q √ 2 a parenrightbigg = 6 k q 2 √ 2 a . The final energy is E f = 1 2 mv 2 . From energy conversation, we have E i = E f 6 k q 2 √ 2 a = 1 2 mv 2 v = q radicalBigg 6 √ 2 k ma . keywords: 002 10.0 points A wire that has a uniform linear charge den sity of 1 . 2 μ C / m is bent into the shape as shown, with radius 2 . 1 m. 4 . 2 m 4 . 2 m P 2 . 1 m Find the electrical potential at point P. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 57579 . 5 V. Explanation: Let : λ = 1 . 2 μ C / m = 1 . 2 × 10 − 6 C / m , R = 2 . 1 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 dq r = k e integraldisplay 3 R R λdx x = k e λ ln x vextendsingle vextendsingle vextendsingle vextendsingle 3 R R = k e λ [ln(3 R ) − ln R ] = k e λ ln 3 . zhou (ejz224) – Homework 3 – shih – (57480) 2 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 λRdθ R = k e λ integraldisplay π dθ = k e λθ vextendsingle vextendsingle vextendsingle vextendsingle π = k e λπ , so the electric potential at P is V p = V right + V left + V semi = 2 k e λ ln 3 + k e λπ = k e λ (2 ln3 + π ) = (8 . 98755 × 10 9 N · m 2 / C 2 ) × (1 . 2 × 10 − 6 C / m) (2 ln 3 + π ) = 57579 . 5 V . 003 10.0 points A charged particle is connected to a string that is is tied to the pivot point P . The particle, string, and pivot point all lie on a horizontal table (consequently the figure below is viewed from above the table). The particle is initially released from rest when the string makes an angle 54 ◦ with a uniform electric field in the horizontal plane (shown in the figure)....
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This note was uploaded on 04/09/2011 for the course PHY 302l taught by Professor Morrison during the Spring '08 term at University of Texas.
 Spring '08
 morrison
 Physics, Charge, Work

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