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hw8_solns - Answer Key Homework 8 David McIntyre This...

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Answer, Key – Homework 8 – David McIntyre 1 This print-out should have 17 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. Chapter 23 problems. 001 (part 1 of 1) 0 points A solid conducting sphere is given a positive charge Q . How is the charge Q distributed in or on the sphere? 1. It is uniformly distributed throughout the sphere. 2. Its density decreases radially outward from the center. 3. It is concentrated at the center of the sphere. 4. Its density increases radially outward from the center. 5. It is uniformly distributed on the surface of the sphere only. correct Explanation: The charge distribution on conductors can only be on the surface, and since on a spherical surface every point is like any other surface point, the charge distribution is uniform. The electric field is normal to the surface of a conductor. The conductor is symmetrical (since it is spherical), so the charge must be uniform. 002 (part 1 of 1) 10 points Three point charges are located at the vertices of an equilateral triangle. The charge at the top vertex of the triangle is given in the figure. The two charges q at the bottom vertices of the triangle are equal. A fourth charge 3 μ C is placed below the triangle on its symmetry- axis, and experiences a zero net force from the other three charges, as shown in the figure below. 7 . 3 m 5 . 1 m - 3 . 8 μ C q q 3 μ C Find q . Correct answer: 0 . 704413 μ C. Explanation: Given : Q 1 = - 3 . 8 μ C , Q 2 = q , Q 3 = q , Q 4 = 3 μ C , a = 7 . 3 m , and d = - 5 . 1 m . a d Q 1 Q 2 Q 3 3 μ C θ The force on Q 4 is due to the Coulomb forces from Q 1 , Q 2 , and Q 3 . Because Q 2 and Q 3 have equal charge, the x -components of their forces cancel out (by symmetry). Thus we only need to consider the y -components of the forces. Coulomb’s law tells us F i = k Q i Q 4 r 2 is the i th force on Q 4 from the i th charge. The forces from Q 2 and Q 3 are equal to each other, and opposite the direction of the force from Q 1 , since otherwise they could not cancel. Total
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Answer, Key – Homework 8 – David McIntyre 2 force on Q 4 is F Q 4 = k Q 1 Q 4 r 2 14 + 2 k Q 2 Q 4 r 2 24 sin θ with r 14 the distance between Q 4 and Q 1 , r 24 = r 34 the distance between Q 4 and either Q 2 or Q 3 , and θ indicated in the sketch above. Remember that this force F Q 4 will be set equal to zero since the problem tells us the forces are in equilibrium. Because Q 1 , Q 2 , and Q 3 form an equilateral triangle, of sides of length a , it can be seen that r 2 14 = ˆ 3 2 a + | d | ! 2 and r 2 24 = a 2 4 + d 2 . Also, sin θ = | d | r 24 = | d | q d 2 + a 2 4 . Our force equation becomes 0 = k Q 4 Q 1 3 2 a + | d | · 2 + 2 Q 2 a 2 4 + d 2 · | d | q a 2 4 + d 2 . Rearranging, we get Q 2 = - Q 1 2 | d | h a 2 4 + d 2 i 3 / 2 h 3 2 a + | d | i 2 = - ( - 3 . 8 μ C) 2 (5 . 1 m) × h (7 . 3 m) 2 4 + ( - 5 . 1 m) 2 i 3 / 2 h 3 2 (7 . 3 m) + (5 . 1 m) i 2 = 0 . 704413 μ C .
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