PHY303LHW2S

# PHY303LHW2S - rosen(arr956 – Homework 02 – Chiu...

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Unformatted text preview: rosen (arr956) – Homework 02 – Chiu – (58295) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 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) − − + + D A C B a O The magnitude of the electric field E O at the center O is given by 1. E O = 3 k e q a 2 2. E O = 1 4 √ 2 k e q a 2 3. E O = 3 √ 2 k e q a 2 4. E O = 1 √ 2 k e q a 2 5. E O = √ 2 k e q a 2 6. E O = 0 7. E O = 4 √ 2 k e q a 2 correct 8. E O = k e q a 2 9. E O = 1 2 √ 2 k e q a 2 10. E O = 2 √ 2 k e 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 e q parenleftbigg a √ 2 parenrightbigg 2 = 2 k e 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 e 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 e q a 2 parenleftbigg − 1 √ 2 ˆ ı + 1 √ 2 ˆ parenrightbigg and vector E B + vector E D = 4 k e q a 2 parenleftbigg − 1 √ 2 ˆ ı − 1 √ 2 ˆ parenrightbigg , so the total is, adding the ˆ ı and ˆ components separately, vector E = 4 k e q a 2 parenleftbigg − 1 √ 2 ˆ ı + 1 √ 2 ˆ − 1 √ 2 ˆ ı − 1 √ 2 ˆ parenrightbigg rosen (arr956) – Homework 02 – Chiu – (58295) 2 = 4 √ 2 k e q a 2 ( − ˆ ı ) = − 4 √ 2 k e q a 2 ˆ ı bardbl E O bardbl = − 4 √ 2 k e q a 2 . 002 (part 2 of 4) 10.0 points Find the direction of the electric field. 1. − vector j 2. vector j 3. 1 √ 2 ( vector i − vector j ) 4. undetermined as vector E = 0 5....
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PHY303LHW2S - rosen(arr956 – Homework 02 – Chiu...

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