If an object has a charge of 32 10 15 c then it must

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If an object has a charge of +3.2 10 -15 C, then it must contain A 10,000 more protons than electrons. B 10,000 more electrons than protons. C 20,000 more protons than electrons. D 10,000 more protons than neutrons. E 20,000 more electrons than protons. Physics 180B — Fall 2017 — TEST 1: FORM D Solutions — page 3 of 4 See Sheet 1, Problems 5, 6(a), 8, and 9. ( ) ( ) ( )( ) N F k r q q 9 10 10 8 10 10 9 8 12 12 12 12 12 6 2 1 2 9 3 2 9 6 $ # # # # $ $ $ = = = = - - - See Sheet 2, Problem 5. Because the direction of F is opposite the direction of E , we know that the charge q experiencing the electrostatic force must be negative . (This eliminates choices B and E.) All we need to do now is determine the magnitude of q : . C F q E q E F 10 1 8 10 10 1 10 10 4 4 8 2 9 3 4 5 5 9 ( # # # # # = = = = = - - - See Sheet 1, Problem 3. , 20 000 = ( ) . . q n e n e q 1 6 10 3 2 10 2 10 19 15 4 ( # # # = + = + = + + = - - So, because the object has a positive charge, it must contain 20,000 more protons than electrons . See Sheet 1, Problem 7. Call upward the positive direction. Then the force that q 1 exerts on q is 1 ( ) ( )( ) k d q q k d q k d q F 3 15 9 15 3 5 2 2 2 2 2 = + = + = + and the force that q 2 exerts on q is ( ) ( )( ) k d q q k d q F 2 3 4 3 2 2 2 2 = = - - So, by superposition, the net electrostatic force on q is 1 k d q k d q k d q k d q F F 3 5 4 3 3 5 4 3 12 11 2 2 2 2 2 2 2 2 2 + = + + - = - = + e e b o o l
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16. The magnitude of the electrostatic force between two given charged particles is F . If one of the charges is decreased by a factor of 16, and the distance between the charges is reduced by a factor of 2, then the magnitude of the resulting electrostatic force will become A F 4 1 B F 8 C F 64 1 D F 8 1 E F 4 17. Consider two source charges, a negative charge -3 Q and a positive charge + Q , fixed on the x - axis, with -3 Q to the left of the origin at position x = - d /2, and with + Q to the right of the origin at position x = + d /2. Determine the electric field at the point x = 3 d /2. A k d Q 2 1 2 , to the right B k d Q 1 4 2 , to the left C k d Q 2 3 2 , to the right D k d Q 1 4 2 , to the right E k d Q 2 1 2 , to the left 18. At a point P in space, the electric field points to the right and has a magnitude of 4 ◊ 10 6 N/C. Compute the electric force that an ion with charge +5 e would feel if it were placed at P. A 2 ◊ 10 7 N, to the left B 2 ◊ 10 -7 N, to the right C 3.2 ◊ 10 -12 N, to the right D 3.2 ◊ 10 -12 N, to the left E 2 ◊ 10 7 N, to the right END OF TEST Physics 180B — Fall 2017 — TEST 1: FORM D Solutions — page 4 of 4 See Sheet 1, Problem 4.
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  • Fall '17
  • Steven Leduc
  • Physics

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