solution_pdf - Version 005/AAABB midterm 01a turner(56725...

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Version 005/AAABB – midterm 01a – turner – (56725) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points 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 81 with a uniform electric field in the horizontal plane (shown in the figure). 81 1 m 143 V / m P 0 . 01 kg 3 μ C initial release ω parallel P Determine the speed of the particle when the string is parallel to the electric field. 1. 0.157043 2. 0.148059 3. 0.0973579 4. 0.337352 5. 0.79873 6. 0.484921 7. 0.201553 8. 0.168027 9. 0.205448 10. 0.269032 Correct answer: 0 . 269032 m / s. Explanation: Let : L = 1 m , q = 3 μ C = 3 × 10 6 C , m = 0 . 01 kg , and θ = 81 . θ L E P m q Let V = 0 at point P . The potential at the initial position is V i = vector E · vectors = E L cos θ , and the potential at the final position is V f = E L . The potential energy U is given by U = qV , so by conservation of energy ( K + U ) i = ( K + U ) f q E L cos θ = 1 2 m v 2 q E L . v = radicalbigg 2 q E L (1 cos θ ) m . Since q E L = (3 × 10 6 C) (143 V / m) (1 m) = 0 . 000429 m 2 · kg / s 2 , then v = radicalBig 2 (0 . 000429 m 2 · kg / s 2 ) × radicalbigg 1 cos 81 0 . 01 kg = 0 . 269032 m / s . 002 10.0 points Consider a square with side a . Four charges q , + q , + q , and q are placed at the corners A , B , C , and D , respectively (see figure). O - D - A + B + C a
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Version 005/AAABB – midterm 01a – turner – (56725) 2 The magnitude of the electric field E O at the center O is given by 1. E O = 2 k q a 2 2. E O = 1 4 2 k q a 2 3. E O = 3 k q a 2 4. E O = 3 2 k q a 2 5. E O = 4 2 k q a 2 correct 6. E O = 1 3 2 k q a 2 7. E O = 1 2 2 k q a 2 8. E O = 1 2 k q a 2 9. E O = k q a 2 10. E O = 2 2 k q a 2 Explanation: The magnitude of each individual field, say the one from the charge at A , follows imme- diately from the formula, noting the distance between each corner and the center is a 2 , therefore E A = k q parenleftbigg a 2 parenrightbigg 2 = 2 k q a 2 However, since not all the forces are collinear (pointing in the same direction) we cannot simply add the magnitudes; we must carry out vector addition. The force is always along the line connect- ing the charge in question and the field point. The direction of the electric field at the field point O is defined to be the direction in which a positive charge would feel a force. Thus we find that the two negative charges yield a field pointing away from them from O and the two positive charges yield forces pointing to- wards them from O. This is summed up in the following figure: E E A + E C E B + E D where we have drawn the resultant electric field vector vector E as well.
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