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solutio234n2_pdf - kinsey(cbk323 oldhomework 02...

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kinsey (cbk323) – oldhomework 02 – Turner – (56725) 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 10.0 points The diagrams below depict three electric field patterns. Some of these patterns are physi- cally impossible. Assume these electric field patterns are due to static electric charges outside the regions shown. (a) (b) (c) Which electrostatic field patterns are phys- ically possible? 1. (a) and (b) 2. (b) and (c) 3. (b) only correct 4. (a) and (c) 5. (a) only 6. (c) only Explanation: Electrostatic lines of force do not intersect one another. Neither do they form a closed circuit (unless there is a changing magnetic field present). 002 (part 1 of 2) 10.0 points Three charges are arranged as shown in the figure. x y 2 . 3 nC 3 . 7 m 0 . 9 nC 0 . 48 nC 1 . 5 m Find the magnitude of the electro- static force on the charge at the origin. The Coulomb cosnstant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 2 . 19648 nN. Explanation: Let : ( x 0 , y 0 ) = (0 m , 0 m) , q 0 = 0 . 9 nC , ( x 1 , y 1 ) = (3 . 7 m , 0 m) , q 1 = 2 . 3 nC , ( x 2 , y 2 ) = (0 m , 1 . 5 m) , and q 2 = 0 . 48 nC . x y q 1 x 1 q 0 q 2 y 2 F θ The force vector F 10 (in the hatwider 10 direction, ˆ ı ) be- tween charge 9 × 10 10 C and 2 . 3 × 10 9 C is vector F 10 = + k e q 0 q 1 x 2 1 ( ˆ ı ) = ( 8 . 98755 × 10 9 N · m 2 / C 2 )
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kinsey (cbk323) – oldhomework 02 – Turner – (56725) 2 × ( 0 . 9 nC) (2 . 3 nC) (3 . 7 m) 2 ˆ ı = (1 . 35897 × 10 9 N)ˆ ı . The plus sign means the force vector F 10 is towards the positive x -axis. The force vector F 20 (in the hatwider 20 direction, +ˆ ) between charge 9 × 10 10 C and 4 . 8 × 10 10 C is vector F 20 = + k e q 0 q 2 y 2 2 (+ˆ ) = (8 . 98755 × 10 9 N · m 2 / C 2 ) × ( 0 . 9 nC) (0 . 48 nC) ( 1 . 5 m) 2 ˆ = ( 1 . 72561 × 10 9 N)ˆ  . The minus sign means the force vector F 20 is towards the negative y -axis. The magni- tude of the total force exerted on the charge 9 × 10 10 C is equal to bardbl vector F bardbl = radicalBig F 2 10 + F 2 20 = bracketleftBig (1 . 35897 × 10 9 N) 2 + ( 1 . 72561 × 10 9 N) 2 bracketrightBig 1 2 = 2 . 19648 × 10 9 N = 2 . 19648 nN . 003 (part 2 of 2) 10.0 points What is the angle θ between the electrostatic force on the charge at the origin and the pos- itive x -axis? Answer in degrees as an angle between 180 and 180 measured from the positive x
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