Homework 3-solutions

# Homework 3-solutions - zhou(ejz224 Homework 3 shih(57480...

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zhou (ejz224) – Homework 3 – shih – (57480) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points 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 m a 2. bardbl vectorv bardbl = q radicalBigg 6 2 k m a correct 3. bardbl vectorv bardbl = q radicalBigg 3 3 k m a 4. bardbl vectorv bardbl = q radicalBigg 6 3 k m a 5. bardbl vectorv bardbl = q radicalBigg 6 6 k m a 6. bardbl vectorv bardbl = q radicalBigg 3 2 k m a 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 m v 2 . From energy conversation, we have E i = E f 6 k q 2 2 a = 1 2 m v 2 v = q radicalBigg 6 2 k m a . keywords: 002 10.0points 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 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 R ) ln R ] = k e λ ln 3 .

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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 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 , so the electric potential at P is V p = V right + V left + V semi = 2 k e λ ln 3 + k e λ π = k e λ (2 ln 3 + π ) = (8 . 98755 × 10 9 N · m 2 / C 2 ) × (1 . 2 × 10 6 C / m) (2 ln 3 + π ) = 57579 . 5 V . 003 10.0points 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). 54 1 . 8 m 146 V / m P 0 . 016 kg 3 μ C initial release ω parallel P Determine the speed of the particle when the string is parallel to the electric field.
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