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Millikan Experiment Worksheet- Answer Key (Front & Back)

Millikan Experiment Worksheet- Answer Key (Front & Back). + + + + + + +. Fe. E. Fg. - - - - - - -. Front Side 1 . F g = 1.9x10 -15 N [D] “stationary” means F e = - F g so F e = 1.9x10 -15 N [U]

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Millikan Experiment Worksheet- Answer Key (Front & Back)

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  1. Millikan Experiment Worksheet- Answer Key (Front & Back) + + + + + + + Fe E Fg - - - - - - -

  2. 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 + + + + + + +

  3. 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 + + + + + + +

  4. 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 + + + + + + +

  5. 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]) = 9.11x10-14N [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] - 4.85x10-14 N [U] = 4.26x10-14N [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 - - - - - - -

  6. 5. a= 12.4m/s2 [D] m = 5.95x10-15 kg a. Fe= ? 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] - 5.83x10-14 N [D] = 1.55x10-14 N [D] + + + + + + + + a E Fe Fg - - - - - - -

  7. 6. a= 8.0m/s2[D] m = 1.5x10-25 kg Fe= ? I have randomly assigned plates here; it could easily have been the other way a< 9.8m/s2means Fe is up Fnet= ma = (1.5x10-15 kg)(8.0 m/s2 [D]) = 1.2x10-14N [D] Fg= mg = (1.5x10-15 kg)(9.8 m/s2 [D]) = 1.5x10-14N [D] Fe =Fnet - Fg = 1.2x10-14N [D] – 1.5x10-14N [D] = - 3x10-15 N [D] or + 3x10-15 N [U] + + + + + + + Fe q a E Fg - - - - - - -

  8. Back Side 1.a) m = 8.22x10-11 kg 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]) = 8.06x10-10N [D] Fe =- Fg = 8.06x10-10 N [U] q= Fe /E = 8.06x10-10 N [U] 4.36x107 N/C [U] = + 1.85x10-17 C - - - - - - - Fe E + Fg + + + + + + +

  9. 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]) = 5.01x10-10N [D] Fe =Fnet- Fg = 5.01x10-10 N [D] - 8.06x10-10N [D] =-3.05x10-10N [D] OR + 3.05x10-10 N [U] - - - - - - - Fe a E + Fg + + + + + + +

  10. 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]) = 1.43x10-9N [D] Fe =Fnet- Fg = 1.43x10-9N [D] - 8.06x10-10N [D] =6.2x10-10N [D] + + + + + + + + E Fg Fe a - - - - - - -

  11. 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]) = 4.09x10-8N [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 + + + + + + +

  12. 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]) = 1.77x10-7N [D] Fe = Fnet- Fg = 1.77x10-7N [D] - 4.09x10-8 N [D] = 1.36x10-7N [D] + + + + + + + + E Fe Fg - - - - - - -

  13. 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 = 1.16x10-9N [D] - 4.09x10-8 N [D] = -3.97x10-8N [D] OR 3.97x10-8N [U] E = Fe /q = 3.97x10-8 N [U] - - - - - - - Fe + a E Fg + + + + + + + +2.56x10-18C = 1.55x10-10N/C [U]

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