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HW2-solutions

# HW2-solutions - HW2 Tsoi(60250 This print-out should have...

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– HW2 – Tsoi – (60250) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 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. + + + D C A B a The magnitude of the electric field at D due to the charges at A , B , and C is given by 1. bardbl vector E bardbl = 2 k q a 2 2. bardbl vector E bardbl = 7 2 k q a 2 3. bardbl vector E bardbl = 5 4 k q a 2 4. bardbl vector E bardbl = 5 2 k q a 2 5. bardbl vector E bardbl = 3 k q a 2 6. bardbl vector E bardbl = 9 4 k q a 2 7. bardbl vector E bardbl = 3 2 k q a 2 correct 8. bardbl vector E bardbl = k q a 2 9. bardbl vector E bardbl = 2 k q a 2 10. bardbl vector E bardbl = 3 4 k q a 2 Explanation: The magnitudes of the electric fields at D due to A and C are E A = E C = k q a 2 since they are at a distance a from d , whereas E B = k q ( a 2) 2 = k q 2 a 2 since B is at a distance 2 a from d . 292 . 53 E C E A E B E A + E C E A + E B + E C + D As for directions, vector E A (downwards since C is positive) and vector E C (right since A is negative) are at right angles and can be added with Pythagoras theorem: bardbl vector E A + vector E C bardbl = k q a 2 1 + 1 = 2 k q a 2 This resultant vector points 45 down to the right. The vector vector E B points at 135 down to the left. The angle between these is 90 , so we apply Pythagoras’ theorem again, now to all three: bardbl ( vector E A + vector E C ) + vector E B bardbl = radicalBigg parenleftBig 2 parenrightBig 2 + parenleftbigg 1 2 parenrightbigg 2 k q a 2 , which means bardbl vector E bardbl = radicalbigg 9 4 k q a 2 = 3 2 k q a 2 . 002 (part 2 of 2) 10.0 points The polar angle of the corresponding electric field vector at D is within the range 1. 270 θ < 315 correct 2. 225 θ < 270 3. 90 θ < 135 4. 315 θ < 360

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– HW2 – Tsoi – (60250) 2 5. 45 θ < 90 6. 0 θ < 45 7. 135 θ < 180 8. 180 θ < 225 Explanation: The x -components of the vectors are E Cx = 0 E Bx = 1 2 k q 2 a 2 E Ax = E A = k q a 2 , and the y -components are E Cy = E C = k q a 2 E By = 1 2 k q 2 a 2 E Ay = 0 . Thus the components of the total vector are E x = 1 2 k q 2 a 2 + k q a 2 = + parenleftBigg 1 2 4 parenrightBigg k q a 2 E y = k q a 2 1 2 k q 2 a 2 = parenleftBigg 1 + 2 4 parenrightBigg k q a 2 Which means the tangent of the angle α be- tween the resultant vector and the horizontal is tan α = parenleftBigg 1 + 2 4 parenrightBigg k q a 2 + parenleftBigg 1 2 4 parenrightBigg k q a 2 = 2 + 1 2 2 1 2 . Therefore the angle is α = 64 . 4712 + n (180 ) We see from the figure that the resultant must point down, so we must have n = 2 to obtain the right range α = 295 . 529 .
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