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oldhomework 08-solutions

# oldhomework 08-solutions - howard(cah3459 – oldhomework...

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howard (cah3459) – oldhomework 08 – Turner – (56705) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 4) 10.0 points Two charges are located in the ( x, y ) plane as shown in the fgure below. The felds pro- duced by these charges are observed at the origin, p = (0 , 0). The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . x y + Q + Q p b a a Use Coulomb’s law to fnd the x -component oF the electric feld at p . 1. E x = 2 k e Qa ( a 2 + b 2 ) 3 / 2 2. E x = 4 k e ( a 2 + b 2 ) 3 / 2 3. E x =0 correct 4. E x = - 4 k e ( a 2 + b 2 ) 3 / 2 5. E x = 2 k e Q a 2 + b 2 6. E x = - 2 k e ( a 2 + b 2 ) 3 / 2 7. E x = - k e ( a 2 + b 2 ) 3 / 2 8. E x = k e ( a 2 + b 2 ) 3 / 2 9. E x = - 2 k e Q a 2 + b 2 Explanation: Let : k e =8 . 98755 × 10 9 N · m 2 / C 2 . + Q 1 + Q 2 p b a a r 1 = ± x 2 1 + y 2 1 = ² a 2 + b 2 . r 2 = ± x 2 2 + y 2 2 = ± ( - a ) 2 + b 2 = ² a 2 + b 2 , so r 2 = r 1 = r. θ θ E 1 E 2 + Q 1 + Q 2 where | sin θ | = b r = b a 2 + b 2 | cos θ | = a r = a a 2 + b 2 . In the x -direction, the contributions From the two charges are E x 1 = - k e (+ Q ) r 2 1 | cos( θ ) | (1) = - k e (+ Q ) ( a 2 + b 2 ) a a 2 + b 2 = - k e ( a 2 + b 2 ) 3 / 2 E x 2 = - k e ( - Q ) r 2 2 | cos( θ ) | (2) = - k e ( - Q ) ( a 2 + b 2 ) a a 2 + b 2 =+ k e ( a 2 + b 2 ) 3 / 2 E x = E x 1 + E x 2 .

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howard (cah3459) – oldhomework 08 – Turner – (56705) 2 002 (part 2 of 4) 10.0 points Let: V = 0 at inFnity. ±ind the electric potential at p. 1. V y = 4 k e Q a 2 + b 2 2. V y = 4 k e Qa a 2 + b 2 3. V y =0 4. V y = - 4 k e a 2 + b 2 5. V y = - 4 k e Q a 2 + b 2 6. V y = 2 k e a 2 + b 2 7. V y =+ 2 k e Q a 2 + b 2 correct 8. V y = - 2 k e Q a 2 + b 2 9. V y = - 2 k e a 2 + b 2 Explanation: The potential for a point charge + Q is V = k e + Q r . ±or the two charges in this problem, we have V 1 = k e + Q a 2 + b 2 . V 2 = k e + Q a 2 + b 2 . V p = V 1 + V 2 = k e a 2 + b 2 [+ Q + (+ Q )] 2 k e Q a 2 + b 2 . 003 (part 3 of 4) 10.0 points ±ind the y -component of the electric Feld at 1. E y 2 k e Q a 2 + b 2 2. E y = - 4 k e Qb ( a 2 + b 2 ) 3 / 2 3. E y = - 2 k e ( a 2 + b 2 ) 3 / 2 4. E y = - k e ( a 2 + b 2 ) 3 / 2 5. E y k e ( a 2 + b 2 ) 3 / 2 6. E y 2 k e ( a 2 + b 2 ) 3 / 2 correct 7. E y 8. E y = - 2 k e Q a 2 + b 2 9. E y 4 k e ( a 2 + b 2 ) 3 / 2 Explanation: r =( x 2 + y 2 ) 1 / 2 a 2 + b 2 ) 1 / 2 and ∂ r ∂y = 1 2 ( y 2 + x 2 ) - 1 / 2 · 2 y ∂ V ∂r = - k e - Q r 2 . To calculate the electric Feld from the poten- tial E y = - = - . So E y = k e - Q r 2 ± 1 2 ( y 2 + x 2 ) - 1 / 2 · 2 y ² = k e - Q r 2 · cos(Δ θ ) (3) = k e - Q r 2 · y r (4) = k e - Qy r 3 2 k e ( a 2 + b 2 ) 3 / 2 . 004 (part 4 of 4) 10.0 points
howard (cah3459) – oldhomework 08 – Turner – (56705) 3 Point p is the origin (0 , 0) and point s is at (0 ,s ). Find V (0 , 0) - V (0 )[ e.g. ,( V p - V s )]. 1. V p - V s = 2 k e Q ± a 2 +( s + b ) 2 - k e Q a 2 + b 2 2. V p - V s = 4 k e Q a 2 + b 2 - 2 k e Q ± a 2 s + b ) 2 3. V p - V s = 2 k e Q a 2 s + b ) 2 - 2 k e Q a 2 + b 2 4. V p - V s =4 k e Q ± a 2 s + b ) 2 5. V p - V s = 2 k e Q a 2 + b 2 - 2 k e Q ± a 2 s + b ) 2 correct 6. V p - V s = 4 k e Q a 2 + b 2 - 2 k e Q a 2 s + b ) 2 7. V p - V s = 4 k e Q ± a 2 s + b ) 2 - 2 k e Q a 2 + b 2 8. V p - V s = 2 k e Q a 2 + b 2 - 2 k e Q a 2 s + b ) 2 9. V p - V s = 2 k e Q ± a 2 s + b ) 2 - 2 k e Q a 2 + b 2 10. V p - V s k e Q a 2 s + b ) 2 Explanation: r 1 = ² x 2 1 + y 2 1 = ² a 2 s + b ) 2 .

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