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Unformatted text preview: Version 044/AACDA midterm 01 Turner (59070) 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 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 The magnitude of the electric field E O at the center O is given by 1. E O = 1 4 2 k q a 2 2. E O = 3 2 k q a 2 3. E O = 3 k q a 2 4. E O = 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 = k q a 2 8. E O = 1 2 k q a 2 9. E O = 2 2 k q a 2 10. E O = 1 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. Since e.g. , vector E A and vector E C are collinear, we can add them up bardbl vector E A + vector E C bardbl = E A + E C = 4 k q a 2 and the same value for E B + E D . These two resultant vectors, however, must be vectori- ally added. The Cartesian components of the two vectors with the origin at O are vector E A + vector E B = 4 k q a 2 parenleftbigg 1 2 + 1 2 parenrightbigg and vector E B + vector E D = 4 k q a 2 parenleftbigg 1 2 1 2 parenrightbigg , so the total is, adding the and components separately, vector E = 4 k q a 2 parenleftbigg 1 2 + 1 2 1 2 1 2 parenrightbigg Version 044/AACDA midterm 01 Turner (59070) 2 = 4 2 k q a 2 ( ) = 4 2 k q a 2 bardbl E O bardbl = 4 2 k q a 2 . 002 10.0 points Consider a long, uniformly charged, cylindri- cal insulator of radius R with charge density 1 . 1 C / m 3 . (The volume of a cylinder with radius r and length is V = r 2 .) The value of the Permittivity of free space is 8 . 85419 10 12 C 2 / N m 2 R 2 . 3 cm What is the magnitude of the electric field inside the insulator at a distance 2 . 3 cm from the axis (2 . 3 cm < R )?...
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
- Spring '08