HW #2-solutions - sobotik (mrs2825) HW #2 Antoniewicz...

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Unformatted text preview: sobotik (mrs2825) HW #2 Antoniewicz (57970) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 4) 3.0 points A charge 1 . 1 10 5 C is fixed on the x-axis at 7 m, and a charge 1 10 5 C is fixed on the y-axis at 6 m, as shown on the diagram. 7 m 1 . 1 10 5 C 6m 1 10 5 C Calculate the magnitude of the resultant electric field vector E at the origin. Correct answer: 3209 . 9 N / C. Explanation: Let : k = 8 . 98755 10 9 N m 2 / C 2 , Q x = 1 . 1 10 5 C , Q y = 1 10 5 C , Q = 3 10 6 C , x = 7 m , and y = 6 m . For charge Q x , | E x | = k | Q x | x 2 = (8 . 98755 10 9 N m 2 / C 2 ) vextendsingle vextendsingle 1 . 1 10 5 C vextendsingle vextendsingle (7 m) 2 = 2017 . 61 N / C . For charge Q y , | E y | = k | Q y | y 2 = (8 . 98755 10 9 N m 2 / C 2 ) 1 10 5 C (6 m) 2 = 2496 . 54 N / C . The resultant electric field at the origin is shown on the diagram. 2017 . 61 N / C 2496 . 54N / C bardbl vector E bardbl Thus, bardbl vector E bardbl = radicalBig E 2 x + E 2 y = radicalBig (2017 . 61 N / C) 2 + (2496 . 54 N / C) 2 = 3209 . 9 N / C . 002 (part 2 of 4) 3.0 points Calculate the electric potential at the origin. Correct answer: 855 . 957 V. Explanation: The electric potential of a point charge is V = k q r , so the electric potential at the origin (due to charges Q x and Q y ) is thus V = k parenleftbigg Q x x + Q y y parenrightbigg = (8 . 98755 10 9 N m 2 / C 2 ) parenleftbigg 1 . 1 10 5 C 7 m + 1 10 5 C 6 m parenrightbigg = 855 . 957 V . 003 (part 3 of 4) 2.0 points A 3 10 6 C charge is brought from a very distant point by an external force and placed at the origin. Calculate the magnitude of the electric force on this charge. sobotik (mrs2825) HW #2 Antoniewicz (57970) 2 Correct answer: 0 . 00962971 N. Explanation: We know that vector F = q vector E . The electric force on Q is shown on the diagram. x y bardbl vector F bardbl bardbl vector F bardbl = | Q | bardbl vector E bardbl = vextendsingle vextendsingle 3 10 6 C vextendsingle vextendsingle (3209 . 9 N / C) = . 00962971 N . 004 (part 4 of 4) 2.0 points Calculate the work that had to be done by an external force to bring Q to the origin from the distant point. Correct answer: . 00256787 J. Explanation: The external force must do work equal in magnitude to that done by the electric field; i.e. , equal to the change in electric potential energy: W = Q V = ( 3 10 6 C) (855 . 957 V 0) = . 00256787 J ....
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HW #2-solutions - sobotik (mrs2825) HW #2 Antoniewicz...

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