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

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Front Side 1. F g = 1.9x N [D] “stationary” means F e = -F g so F e = 1.9x N [U] E = 6.0x10 3 N/C (if top plate is “-” then [U]) a. q = F e /E = 1.9x N [U] 6.0x10 3 N/C [U] =+ 3.2x C b. If the top plate is “-” then only a “+” charge would be attracted #e lost = 3.2x C 1.6x C = 2 electrons lost E Fe Fg

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2. m = 6.5x kg E = 4.0x10 6 N/C (top plate must be “-” to suspend a “+” drop) a. “suspended” means F e = -F g so F e = 1.9x N [U] q = F e /E = 1.9x N [U] 6.0x10 3 N/C [U] =+ 3.2x C b. If the top plate is “-” then only a “+” charge would be attracted #e lost = 3.2x C 1.6x C = 2 electrons lost E Fe Fg

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3.Has 4 protons so q=(+4)(1.6x C) = +6.4x C (top plate must be “-” to suspend a “+” drop) F g = 6.4x N [D] “suspended” means F e = -F g so F e = 6.4x N [U] E = ? E = F e / q = 6.4x N [U] +6.4x C = 1.0x10 6 N/C [U] E Fe Fg

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4.a= 18.4m/s 2 [D] m = 4.95x kg a. F e = ? a. a> 9.8 m/s 2 means F e is down F net = ma = (4.95x kg)(18.4 m/s 2 [D]) = 9.11x N [D] F g = mg = (4.95x kg)(9.8 m/s 2 [D]) = 4.85x N [D] so F e = F net – F g = 9.11x N [D] x N [U] = 4.26x 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 E Fe Fg a

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5.a= 12.4m/s 2 [D] m = 5.95x kg a. F e = ? b. If top plate is “+” what kind of charge is it? - Positive charge again a. a> 9.8 m/s 2 means F e is down F net = ma = (5.95x kg)(12.4 m/s 2 [D]) = 7.38x N [D] F g = mg = (5.95x kg)(9.8 m/s 2 [D]) = 5.83x N [D] F e = F net - F g = 7.38x N [D] x N [D] = 1.55x N [D] E Fe Fg a

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6.a= 8.0m/s 2 [D] m = 1.5x kg F e = ? I have randomly assigned plates here; it could easily have been the other way a< 9.8m/s 2 means F e is up F net = ma = (1.5x kg)(8.0 m/s 2 [D]) = 1.2x N [D] F g = mg = (1.5x kg)(9.8 m/s 2 [D]) = 1.5x N [D] F e = F net - F g = 1.2x N [D] – 1.5x N [D] = - 3x N [D] or + 3x N [U] q E Fe Fg a

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Back Side 1.a) m = 8.22x kg E = 4.36x10 7 N/C for a positively charged oil drop to be “balanced” means F e = -F g and that the top plate must be negative F g = mg = (8.22x kg)(9.8 m/s 2 [D]) = 8.06x N [D] F e = - F g = 8.06x N [U] q= F e /E = 8.06x N [U] 4.36x10 7 N/C [U] = x C E Fe Fg

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Back Side 1.b) same oildrop so m = 8.22x kg, F g = 8.06x N [D] but a=6.1m/s 2 [D] F e = ? F net = ma = (8.22x kg)(6.1 m/s 2 [D]) = 5.01x N [D] F e = F net - F g = 5.01x N [D] x N [D] = -3.05x N [D] OR x N [U] E Fe Fg a

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Back Side 1.c) same oildrop so m = 8.22x kg, F g = 8.06x N [D] but a=17.4m/s 2 [D] F e = ? F net = ma = (8.22x kg)(17.4 m/s 2 [D]) = 1.43x10 -9 N [D] F e = F net - F g = 1.43x10 -9 N [D] x N [D] = 6.2x N [D] E Fe Fg a

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Back Side 2.a) m = 4.17x10 -9 kg and an excess of 16 protons (lost 16 electrons) “balanced” therefore F e = - F g E = ? q=(+16)(1.6x C) q = +2.56x C F g = mg = (4.17x10 -9 kg)(9.8 m/s 2 [D]) = 4.09x10 -8 N [D] F e = 4.09x10 -8 N [U] E = F e /q = 4.09x10 -8 N [U] +2.56x C = 1.60x10 10 N/C E Fe Fg

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

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

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