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E (µN/C) 2 4.5 6.5 x (nm) -4 Figure 1: 1. The electric field, E, as a function of x for a certain charge distribution is shown in figure 1. (a) The force that a positive point charge, q = 1.6 x 10-1ºC, experiences at the following positions, (i)x = 0 , (ii)x = 1nm and (iii)x = 4nm (i) F1 = 0 (True,False) (ii) F2 = (2µN/C)(1.6 × 10-1ºC) = 3.2 × 10¬25 N (to the left) (True,False) (iii) F3 = (4µN/C)(1.6 × 10-1ºC) = 6.4 × 10–25 N (to the left) (True,False) (b) Determine the work done on this positive point charge between (i) x = 0 to x = 2nm (ii)x = 2nm to x = 4.5nm (iii)x = 4.5nm to x = 6.5nm (iv)x = 0 to x = 6.5nm (i)W1 = |Ē - dĩ = q (1.6 × 10¬19C)(- (-4µN/C)x (2nm) = -6.4 x 10-34 j • dř = q Edх — q хаrea %3 (True,False) - S (ii)W2 = F. dr = TĒ : dĩ = q Edx = q × area = (1.6 × 10¬19C)((-4µN/C) × ((4.5 – 2)nm) = -16 x 10-34 J (True,False) (iii)W3 = | F - dř = q |Ē · dř = q S Edx (1.6 x 10¬19C)(-4µN/C)x((6.5–4.5)nm) = 4 = -6.4 × 10-34 j (True,False) X area = (iv) W4 = -28.8 x 10-34 J (True,False) (c) Determine the work done per unit positive point charge for the above intervals[potential differences]: W1 (i)Vo – V2 = = -4 x 10-15J/C = -4 × 10-15Volts (True,False) W2 (ii)V2 – V4.5 = = -10 x 10-15 J/C = –10 × 10-15Volts (True,False) W3 (iii)V4.5 – V6.5 = = -4 x 10-15 J/C = -4 × 10-15Volts (True,False) W4 (iv)Vo – V6.5 = -4 x 10-15 J/C = –18 × 10-15Volts (True,False)

E (µN/C) 2 4.5 6.5 x (nm) -4 Figure 1: 1. The electric field, E, as a function of x for a certain charge distribution is shown in figure 1. (a) The force that a positive point charge, q = 1.6 x 10-1ºC, experiences at the following positions, (i)x = 0 , (ii)x = 1nm and (iii)x = 4nm (i) F1 = 0 (True,False) (ii) F2 = (2µN/C)(1.6 × 10-1ºC) = 3.2 × 10¬25 N (to the left) (True,False) (iii) F3 = (4µN/C)(1.6 × 10-1ºC) = 6.4 × 10–25 N (to the left) (True,False) (b) Determine the work done on this positive point charge between (i) x = 0 to x = 2nm (ii)x = 2nm to x = 4.5nm (iii)x = 4.5nm to x = 6.5nm (iv)x = 0 to x = 6.5nm (i)W1 = |Ē - dĩ = q (1.6 × 10¬19C)(- (-4µN/C)x (2nm) = -6.4 x 10-34 j • dř = q Edх — q хаrea %3 (True,False) - S (ii)W2 = F. dr = TĒ : dĩ = q Edx = q × area = (1.6 × 10¬19C)((-4µN/C) × ((4.5 – 2)nm) = -16 x 10-34 J (True,False) (iii)W3 = | F - dř = q |Ē · dř = q S Edx (1.6 x 10¬19C)(-4µN/C)x((6.5–4.5)nm) = 4 = -6.4 × 10-34 j (True,False) X area = (iv) W4 = -28.8 x 10-34 J (True,False) (c) Determine the work done per unit positive point charge for the above intervals[potential differences]: W1 (i)Vo – V2 = = -4 x 10-15J/C = -4 × 10-15Volts (True,False) W2 (ii)V2 – V4.5 = = -10 x 10-15 J/C = –10 × 10-15Volts (True,False) W3 (iii)V4.5 – V6.5 = = -4 x 10-15 J/C = -4 × 10-15Volts (True,False) W4 (iv)Vo – V6.5 = -4 x 10-15 J/C = –18 × 10-15Volts (True,False)

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E (µN/C) 2 4.5 6.5 x (nm) -4 Figure 1: 1. The electric field, E, as a function of x for a certain charge distribution is shown in figure 1. (a) The force that a positive point charge, q = 1.6 x 10-1ºC, experiences at the following positions, (i)x = 0 , (ii)x = 1nm and (iii)x = 4nm (i) F1 = 0 (True,False) (ii) F2 = (2µN/C)(1.6 × 10-1ºC) = 3.2 × 10¬25 N (to the left) (True,False) (iii) F3 = (4µN/C)(1.6 × 10-1ºC) = 6.4 × 10–25 N (to the left) (True,False) (b) Determine the work done on this positive point charge between (i) x = 0 to x = 2nm (ii)x = 2nm to x = 4.5nm (iii)x = 4.5nm to x = 6.5nm (iv)x = 0 to x = 6.5nm (i)W1 = |Ē - dĩ = q (1.6 × 10¬19C)(- (-4µN/C)x (2nm) = -6.4 x 10-34 j • dř = q Edх — q хаrea %3 (True,False) - S (ii)W2 = F. dr = TĒ : dĩ = q Edx = q × area = (1.6 × 10¬19C)((-4µN/C) × ((4.5 – 2)nm) = -16 x 10-34 J (True,False) (iii)W3 = | F - dř = q |Ē · dř = q S Edx (1.6 x 10¬19C)(-4µN/C)x((6.5–4.5)nm) = 4 = -6.4 × 10-34 j (True,False) X area = (iv) W4 = -28.8 x 10-34 J (True,False) (c) Determine the work done per unit positive point charge for the above intervals[potential differences]: W1 (i)Vo – V2 = = -4 x 10-15J/C = -4 × 10-15Volts (True,False) W2 (ii)V2 – V4.5 = = -10 x 10-15 J/C = –10 × 10-15Volts (True,False) W3 (iii)V4.5 – V6.5 = = -4 x 10-15 J/C = -4 × 10-15Volts (True,False) W4 (iv)Vo – V6.5 = -4 x 10-15 J/C = –18 × 10-15Volts (True,False)
Transcribed Image Text:E (µN/C) 2 4.5 6.5 x (nm) -4 Figure 1: 1. The electric field, E, as a function of x for a certain charge distribution is shown in figure 1. (a) The force that a positive point charge, q = 1.6 x 10-1ºC, experiences at the following positions, (i)x = 0 , (ii)x = 1nm and (iii)x = 4nm (i) F1 = 0 (True,False) (ii) F2 = (2µN/C)(1.6 × 10-1ºC) = 3.2 × 10¬25 N (to the left) (True,False) (iii) F3 = (4µN/C)(1.6 × 10-1ºC) = 6.4 × 10–25 N (to the left) (True,False) (b) Determine the work done on this positive point charge between (i) x = 0 to x = 2nm (ii)x = 2nm to x = 4.5nm (iii)x = 4.5nm to x = 6.5nm (iv)x = 0 to x = 6.5nm (i)W1 = |Ē - dĩ = q (1.6 × 10¬19C)(- (-4µN/C)x (2nm) = -6.4 x 10-34 j • dř = q Edх — q хаrea %3 (True,False) - S (ii)W2 = F. dr = TĒ : dĩ = q Edx = q × area = (1.6 × 10¬19C)((-4µN/C) × ((4.5 – 2)nm) = -16 x 10-34 J (True,False) (iii)W3 = | F - dř = q |Ē · dř = q S Edx (1.6 x 10¬19C)(-4µN/C)x((6.5–4.5)nm) = 4 = -6.4 × 10-34 j (True,False) X area = (iv) W4 = -28.8 x 10-34 J (True,False) (c) Determine the work done per unit positive point charge for the above intervals[potential differences]: W1 (i)Vo – V2 = = -4 x 10-15J/C = -4 × 10-15Volts (True,False) W2 (ii)V2 – V4.5 = = -10 x 10-15 J/C = –10 × 10-15Volts (True,False) W3 (iii)V4.5 – V6.5 = = -4 x 10-15 J/C = -4 × 10-15Volts (True,False) W4 (iv)Vo – V6.5 = -4 x 10-15 J/C = –18 × 10-15Volts (True,False)
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