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14-2 - brown(twb493 Ch14-h2 chiu(56565 1 This print-out...

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brown (twb493) – Ch14-h2 – chiu – (56565) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points 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 = ( 0 . 7 m , 0 , 0 ) , the electric field will be of the form vector E = ( E x , 0 , 0 ) . Find E x . Correct answer: 128 . 394 N / C. Explanation: The general equation for the electric field at P is vector E = 1 4 πǫ 0 q | vectorr | 2 ˆ r. The x component would be given by E x = 1 4 πǫ 0 q | vectorr | 2 ˆ r x . We have the various quantities that we need: q = 7 nC = 7 × 10 - 9 C | vectorr | = 0 . 7 m ˆ r x = 1 . Plugging in, we obtain E x = 1 4 πǫ 0 7 × 10 - 9 C (0 . 7 m) 2 = 128 . 394 N / C . 002(part1of2)10.0points You want to create an electric field vector E = ( 0 , 3979 N / C , 0 ) 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 πǫ 0 q | vectorr | 2 ˆ r. In this case, q = 1 . 6 × 10 - 19 C for a proton, and ˆ r = ( 0 , 1 , 0 ) , 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 πǫ 0 1 . 6 × 10 - 19 C | vectorr | 2 ⇒ | vectorr | 2 = 1 4 πǫ 0 1 . 6 × 10 - 19 C 3979 N / C ⇒ | vectorr | = ± radicalBigg 1 4 πǫ 0 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(part2of2)10.0points 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 electron’s 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.

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brown (twb493) – Ch14-h2 – chiu – (56565) 2 004(part1of6)10.0points A π - (“pi-minus”) particle, which has charge e , is at location vectorr π = ( 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 π - 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. Explanation: Let’s write the general expression for the electric field at P due to the π - as vector E = 1 4 πǫ 0 Q π - | vectorr | 2 ˆ r . Therefore in order to determine the com- ponents of the field, we need to find the unit vector ˆ r pointing from the location of the π - to the point P and the distance between those two locations, vextendsingle vextendsingle vextendsingle vectorr vextendsingle vextendsingle vextendsingle .
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