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**Millikan Experiment Worksheet- Answer Key (Front & Back)**

Fe E Fg

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Front Side 1. Fg = 1.9x10-15 N [D] “stationary” means Fe= -Fg so Fe = 1.9x10-15 N [U] E = 6.0x103 N/C (if top plate is “-” then [U]) a. q = Fe /E = 1.9x10-15 N [U] 6.0x103 N/C [U] =+ 3.2x10-19 C b. If the top plate is “-” then only a “+” charge would be attracted #e lost = 3.2x10-19 C 1.6x10-19 C = 2 electrons lost Fe + E Fg

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2. m = 6.5x10-14 kg E = 4.0x106 N/C (top plate must be “-” to suspend a “+” drop) a. “suspended” means Fe= -Fg so Fe = 1.9x10-15 N [U] q = Fe /E = 1.9x10-15 N [U] 6.0x103 N/C [U] =+ 3.2x10-19 C b. If the top plate is “-” then only a “+” charge would be attracted #e lost = 3.2x10-19 C 1.6x10-19 C = 2 electrons lost Fe + E Fg

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**E + Fe Fg Has 4 protons so q=(+4)(1.6x10-19C) = +6.4x10-19C**

(top plate must be “-” to suspend a “+” drop) Fg = 6.4x10-13 N [D] “suspended” means Fe= -Fg so Fe = 6.4x10-13 N [U] E = ? E = Fe /q = 6.4x10-13 N [U] +6.4x10-19C = 1.0x106 N/C [U] Fe E + Fg

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**+ E Fe Fg a a= 18.4m/s2 [D] m = 4.95x10-15 kg**

a. Fe= ? a. a> 9.8 m/s2 means Fe is down Fnet= ma = (4.95x10-15 kg)(18.4 m/s2 [D]) = 9.11x10-14 N [D] Fg = mg = (4.95x10-15 kg)(9.8 m/s2 [D]) = 4.85x10-14 N [D] so Fe = Fnet – Fg = 9.11x10-14 N [D] x10-14 N [U] = 4.26x10-14 N [D] b. If top plate is “+”, what kind of charge is it? Positive charge would be repelled by a positive plate/attracted by a negative plate. + Fe Fg a E

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**+ E a Fe Fg 5. a= 12.4m/s2 [D] m = 5.95x10-15 kg = 1.55x10-14 N [D]**

b. If top plate is “+” what kind of charge is it? - Positive charge again a. a> 9.8 m/s2 means Fe is down Fnet= ma = (5.95x10-15 kg)(12.4 m/s2 [D]) = 7.38x10-14 N [D] Fg = mg = (5.95x10-15 kg)(9.8 m/s2 [D]) = 5.83x10-14 N [D] Fe = Fnet - Fg = 7.38x10-14 N [D] x10-14 N [D] = 1.55x10-14 N [D] + a E Fe Fg

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**E Fe q a Fg 6. a= 8.0m/s2 [D] m = 1.5x10-25 kg**

I have randomly assigned plates here; it could easily have been the other way a< 9.8m/s2 means Fe is up Fnet= ma = (1.5x10-15 kg)(8.0 m/s2 [D]) = 1.2x10-14 N [D] Fg = mg = (1.5x10-15 kg)(9.8 m/s2 [D]) = 1.5x10-14 N [D] Fe = Fnet - Fg = 1.2x10-14 N [D] – 1.5x10-14 N [D] = - 3x10-15 N [D] or + 3x10-15 N [U] Fe q a E Fg

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**E + Fe Fg Back Side 1.a) m = 8.22x10-11 kg Fg = mg q= Fe /E**

E = 4.36x107 N/C for a positively charged oil drop to be “balanced” means Fe= -Fg and that the top plate must be negative Fg = mg = (8.22x10-11 kg)(9.8 m/s2 [D]) = 8.06x10-10 N [D] Fe = - Fg = 8.06x10-10 N [U] q= Fe /E 4.36x107 N/C [U] = x10-17 C Fe E + Fg

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**E + Fe a Fg Back Side 1.b) same oildrop so**

m = 8.22x10-11 kg, Fg = 8.06x10-10 N [D] but a=6.1m/s2 [D] Fe= ? Fnet = ma = (8.22x10-11 kg)(6.1 m/s2 [D]) = 5.01x10-10 N [D] Fe = Fnet - Fg = 5.01x10-10 N [D] x10-10 N [D] = -3.05x10-10 N [D] OR x10-10 N [U] Fe a E + Fg

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**+ E Fg Fe a Back Side 1.c) same oildrop so**

m = 8.22x10-11 kg, Fg = 8.06x10-10 N [D] but a=17.4m/s2 [D] Fe= ? Fnet = ma = (8.22x10-11 kg)(17.4 m/s2 [D]) = 1.43x10-9 N [D] Fe = Fnet - Fg = 1.43x10-9 N [D] x10-10 N [D] = 6.2x10-10 N [D] + E Fg Fe a

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**+ E Fe Fg “balanced” therefore Fe = - Fg E= ? q=(+16)(1.6x10-19C)**

Back Side 2.a) m = 4.17x10-9 kg and an excess of 16 protons (lost 16 electrons) “balanced” therefore Fe = - Fg E= ? q=(+16)(1.6x10-19C) q = +2.56x10-18C Fg = mg = (4.17x10-9 kg)(9.8 m/s2 [D]) = 4.09x10-8 N [D] Fe = 4.09x10-8 N [U] E = Fe /q = 4.09x10-8 N [U] +2.56x10-18C = 1.60x1010 N/C Fe + E Fg

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**+ E Fe Fg but a=42.5m/s2 [D] = (4.17x10-9 kg)(42.5 m/s2 [D])**

Back Side 2.b) same drop so m = 4.17x10-9 kg and Fg = 4.09x10-8 N [D] but a=42.5m/s2 [D] Fe = ? Fnet = ma = (4.17x10-9 kg)(42.5 m/s2 [D]) = 1.77x10-7 N [D] Fe = Fnet - Fg = 1.77x10-7 N [D] x10-8 N [D] = 1.36x10-7 N [D] + E Fe Fg

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**+ E Fe a Fg but a=0.279m/s2 [D] = (4.17x10-9 kg)(0.279 m/s2 [D])**

Back Side 2.c) same drop so m = 4.17x10-9 kg and Fg = 4.09x10-8 N [D] but a=0.279m/s2 [D] E = ? Fnet = ma = (4.17x10-9 kg)(0.279 m/s2 [D]) = 1.16x10-9 N [D] Fe = Fnet - Fg = 1.16x10-9 N [D] x10-8 N [D] = x10-8 N [D] OR x10-8 N [U] E = Fe /q = 3.97x10-8 N [U] Fe + a E Fg +2.56x10-18C = 1.55x10-10 N/C [U]

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