oldhomework 08-solutions

oldhomework 08-solutions - howard (cah3459) oldhomework 08...

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Unformatted text preview: 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 find 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 figure below. The fields 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 Coulombs law to find the x-component of the electric field at p . 1. E x = 2 k e Q a ( a 2 + b 2 ) 3 / 2 2. E x = 4 k e Q a ( a 2 + b 2 ) 3 / 2 3. E x = 0 correct 4. E x =- 4 k e Q a ( a 2 + b 2 ) 3 / 2 5. E x = 2 k e Q a 2 + b 2 6. E x =- 2 k e Q a ( a 2 + b 2 ) 3 / 2 7. E x =- k e Q a ( a 2 + b 2 ) 3 / 2 8. E x = k e Q a ( 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 = a 2 + b 2 . r 2 = x 2 2 + y 2 2 = (- a ) 2 + b 2 = a 2 + b 2 , so r 2 = r 1 = r . E 1 E 2 + Q 1 + Q 2 where | sin | = b r = b a 2 + b 2 | cos | = 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( ) | (1) =- k e (+ Q ) ( a 2 + b 2 ) a a 2 + b 2 =- k e Q a ( a 2 + b 2 ) 3 / 2 E x 2 =- k e (- Q ) r 2 2 | cos( ) | (2) =- k e (- Q ) ( a 2 + b 2 ) a a 2 + b 2 = + k e Q a ( a 2 + b 2 ) 3 / 2 E x = E x 1 + E x 2 = 0 . howard (cah3459) oldhomework 08 Turner (56705) 2 002 (part 2 of 4) 10.0 points Let: V = 0 at infinity. Find the electric potential at p. 1. V y = 4 k e Q a 2 + b 2 2. V y = 4 k e Q a a 2 + b 2 3. V y = 0 4. V y =- 4 k e Q a a 2 + b 2 5. V y =- 4 k e Q a 2 + b 2 6. V y = 2 k e Q a 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 Q a a 2 + b 2 Explanation: The potential for a point charge + Q is V = k e + Q r . For 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 Find the y-component of the electric field at p. 1. E y = + 2 k e Q a 2 + b 2 2. E y =- 4 k e Q b ( a 2 + b 2 ) 3 / 2 3. E y =- 2 k e Q b ( a 2 + b 2 ) 3 / 2 4. E y =- k e Q b ( a 2 + b 2 ) 3 / 2 5. E y = + k e Q b ( a 2 + b 2 ) 3 / 2 6. E y = + 2 k e Q b ( a 2 + b 2 ) 3 / 2 correct 7. E y = 0 8. E y =- 2 k e Q a 2 + b 2 9. E y = + 4 k e Q b ( 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 field from the poten- tial E y =- V y =- V r r y ....
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This note was uploaded on 09/15/2010 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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oldhomework 08-solutions - howard (cah3459) oldhomework 08...

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