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Unformatted text preview: brown (twb493) Ch14h2 chiu (56565) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A particle with charge +7 nC (a nanocoulomb is 1 10 9 C) is located at the origin. At a point P given by vectorr = ( . 7 m , , ) , the electric field will be of the form vector E = ( E x , , ) . Find E x . Correct answer: 128 . 394 N / C. Explanation: The general equation for the electric field at P is vector E = 1 4 q  vectorr  2 r. The x component would be given by E x = 1 4 q  vectorr  2 r x . We have the various quantities that we need: q = 7 nC = 7 10 9 C  vectorr  = 0 . 7 m r x = 1 . Plugging in, we obtain E x = 1 4 7 10 9 C (0 . 7 m) 2 = 128 . 394 N / C . 002 (part 1 of 2) 10.0 points You want to create an electric field vector E = ( , 3979 N / C , ) at the origin. Find the y coordinate where you would need to place a proton, in order to create this field at the origin. Correct answer: 6 . 01165 10 7 m. Explanation: The equation for the electric field, in gen eral, is vector E = 1 4 q  vectorr  2 r. In this case, q = 1 . 6 10 19 C for a proton, and r = ( , 1 , ) , since the field points in the positive y direc tion.  vectorr  is the unknown we want to solve for. We can write E y = 3979 N / C = 1 4 1 . 6 10 19 C  vectorr  2  vectorr  2 = 1 4 1 . 6 10 19 C 3979 N / C  vectorr  = radicalBigg 1 4 1 . 6 10 19 C 3979 N / C = 6 . 01165 10 7 m . We choose the minus sign to place the pro ton below the origin; this way the field will point upward like we want it to. 003 (part 2 of 2) 10.0 points Instead of a proton, where would you place an electron to produce the same field at the origin? Correct answer: 6 . 01165 10 7 m. Explanation: This is simple. The proton and electron carry the same magnitude of charge, but the electrons charge is negative. The calculation from above will be exactly the same, but we choose a positive sign at the end to place the electron above the origin. This way the field will still point upward as we want it to. brown (twb493) Ch14h2 chiu (56565) 2 004 (part 1 of 6) 10.0 points A  (piminus) particle, which has charge e , is at location vectorr = ( 8 10 9 , 3 10 9 , 5 10 9 ) m . You will find the components of the electric field at another point P given by vectorr P = ( 7 10 9 , 7 10 9 , 4 10 9 ) m due to the  particle. Begin by finding the x component of the field, which we will call E x ( P ). Answer in N/C. Correct answer: 2 . 63672 10 6 N / C....
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This note was uploaded on 12/01/2011 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics

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