Ch22-h2-extra-solutions.pdf

# Ch22-h2-extra-solutions.pdf - agha(msa2448 Ch22-h2-extra...

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agha (msa2448) – Ch22-h2-extra – turner – (90130) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points What is the direction, if any, of the electric field due to this configuration at a point X >> l >> s ? 1. y 2. + x 3. + y correct 4. vector E ( X ) = 0 5. x Explanation: Since the dipole to the right of the origin is closer to point X , its electric field will dom- inate; since the moment of the dipole points in the − − y direction, the electric field at X points in the + y direction. 002(part2of2)10.0points Using a procedure similar to that used to calculate the electric field of a dipole, find an approximate algebraic expression for | vector E ( X ) | , the magnitude of the electric field due to the configuration at this point X , X >> l >> s . 1. 3 k | vectorp | l X 3 2. 0 3. 3 k | vectorp | s X 3 4. 6 k | vectorp | l X 4 5. 6 k | vectorp | s X 4 6. 6 k | vectorp | X 3 7. 3 k | vectorp | X 4 8. k | vectorp | s X 4 9. 3 k | vectorp | l X 4 correct 10. 3 k | vectorp | ls X 5 Explanation: Using the formula for the electric field of a dipole along its perpendicular axis, vector E dip = k vectorp r 3 and expressing the distance to the left and right dipoles as X + l/ 2 and X l/ 2 respec- tively, we apply the superposition principle at X: vector E net ( X ) = k | vectorp | ( X + l/ 2) 3 ( ˆ y ) + k | vectorp | ( X l/ 2) 3 ˆ y vector E net ( X ) = k | vectorp | parenleftbigg 1 ( X l/ 2) 3 1 ( X + l/ 2) 3 parenrightbigg ˆ y vector E net ( X ) = k | vectorp | X 3 parenleftBigg parenleftbigg 1 l 2 X parenrightbigg 3 parenleftbigg 1 + l 2 X parenrightbigg 3 parenrightBigg ˆ y Making the small argument approximation (since l << X ) and taking the magnitude we obtain an approximate value for | vector E net ( X ) | : k | vectorp | X 3 parenleftbiggparenleftbigg 1 3 parenleftbigg l 2 X parenrightbiggparenrightbigg parenleftbigg 1 3 parenleftbigg l 2 X parenrightbiggparenrightbiggparenrightbigg therefore, vector E net ( X ) k | vectorp | X 3 6 l 2 X vector E net ( X ) 3 k | vectorp | l X 4 003 10.0points

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agha (msa2448) – Ch22-h2-extra – turner – (90130) 2 Consider the setup shown in the figure, where charges Q1, Q2 and Q3 and the point A occupy four corners of a square with the length a at each side. Given that Q 1 = Q 3 = q > 0 and Q 2 = 2 q . The resultant electric field at A contributed by charges Q1 and Q3 is labeled as vector E 13 . The electric field at A contributed by Q2 is la- beled as vector E 2 . Verify that the vectors vector E 13 and vector E 2 are aligned along a same line. Such an alignment implies that the ratio vector E 13 vector E 2 may be represented by a number. Here, a positive ra- tio implies that two vectors are pointing in the same direction and a negative ratio im- plies that they are pointing in the opposite direction. Determine this ratio.
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