HW #2-solutions - sobotik(mrs2825 HW#2 Antoniewicz(57970...

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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 – fnd all choices beFore answering. 001 (part 1 oF 4) 3.0 points A charge 1 . 1 × 10 5 C is fxed on the x -axis at 7 m, and a charge 1 × 10 5 C is fxed on the y -axis at 6 m, as shown on the diagram. 7 m 1 . 1 × 10 5 C 6 m 1 × 10 5 C Calculate the magnitude oF the resultant electric feld v 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 . ±or charge Q x , | E x | = k | Q x | x 2 = (8 . 98755 × 10 9 N · m 2 / C 2 ) × v v 1 . 1 × 10 5 C v v (7 m) 2 = 2017 . 61 N / C . ±or 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 feld at the origin is shown on the diagram. 2017 . 61 N / C 2496 . 54 N / b v E Thus, b v E b = r E 2 x + E 2 y = r (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 p Q x x + Q y y P = (8 . 98755 × 10 9 N · m 2 / C 2 ) × p 1 . 1 × 10 5 C 7 m + 1 × 10 5 C 6 m P = 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.
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sobotik (mrs2825) – HW #2 – Antoniewicz – (57970) 2 Correct answer: 0 . 00962971 N. Explanation: We know that v F = q v E . The electric force on Q is shown on the diagram. x y b v F b v F b = | Q | b v E b = v v 3 × 10 6 C v v (3209 . 9 N / C) = 0 . 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: 0 . 00256787 J. Explanation: The external force must do work equal in magnitude to that done by the electric Feld; i.e. , equal to the change in electric potential energy: W = Q Δ V = ( 3 × 10 6 C) (855 . 957 V 0) = 0 . 00256787 J . 005 10.0 points Two charges, 2 Q and + Q , are located on the x -axis, as shown below. Point P , at a distance of 3 D from the origin O , is one of two points on the positive x -axis at which the electric potential is zero. x 0 2 Q + Q P D D D How far d from the origin O is the other point? 1. d = 6 5 D 2. d = 5 3 D correct 3. d = 8 5 D 4. d = 7 5 D 5. d = 5 4 D 6. d = 4 3 D 7. d = 3 2 D 8. None of these 9. d = 7 4 D 10. d = 9 5 D Explanation: Suppose the zero-potential point is at a distance d from O. The electric potentials are 2 Q 4 π ǫ 0 | D d | and Q 4 π ǫ 0 | 2 D d | , respectively. ±or zero electric potential 2 Q 4 π ǫ 0 | D d | + Q 4 π ǫ 0 | 2 D d | = 0 , 2 | D d | = 1 | 2 D d | , which gives d = 5 3 D or d = 3 D .
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HW #2-solutions - sobotik(mrs2825 HW#2 Antoniewicz(57970...

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