Calculate the charge of the oil drop

Millikan's Oil Drop Experiment Purpose To determine the number of excess electrons on an oil drop, based on the time it takes to rise and fall in the region between charged plates. To verify the charge of a single electron as compared to its known, accepted value. Theory When a small droplet of oil falls through the air, it quickly reaches a very slow terminal velocity when the drag force becomes equal to the force of gravity. V mg Stoke's Law can be used to determine the drag force on a spherical object moving through a fluid. When the sphere is small, and the speed is very slow: F, = kv = 6arv F, = mg & mg = (pV)g Fr = kvfall = 6unrvfall = pVg Ganrv fall = P (3073) g Isolating the expression above for "r" gives the following, which we will refer to as equation #1: Invfall r= 2pg When the oil droplet rises under the influence of an electric field, the free-body diagram becomes: AqE VF . mg V qE = F, + mg qE = kvrise + mg mg qE = Vfall Vrise + mg qE = mg Brisk + 1) Isolating the expression above for "q" gives the following, which we will refer to as equation #2: mg q = Eif only physics were so simple, and Stoke's Law held perfectly true without exception, then finding the charge of the oil drop would be a fairly straightforward two step process: 1) use equation #1 find the radius of the oil drop based on the terminal velocity while falling, then 2) find the charge using equation #2, based on the upward motion under the influence of the electric field. Invfall r= equation 1 2pg equation 2 The problem is that Stoke's Law is not perfect for oil droplets of such a small size. The calculations can be corrected by replacing the viscosity of air n with an effective viscosity neff. neff = 1 equation 3 What a mess! Equation #1 tells us that the radius of the drop is a function of viscosity, but equation #3 in turn says viscosity is a function of radius. If we put it together we can make one corrected equation to solve for "r" (as long as you are not afraid of applying the quadratic formula). Start by squaring both sides of equation #1... 12 = 9nefffall 9vfall 2pg 2pg b invfall pr 2pg b Invfall = 0 2pg 2 367Vfall 2pg r= 2 b 9nvfall r= 2p 2pg + 2p equation 4 m = pV = p_mr3 equation 5 E = AY equation 6 equation 7 r - The radius of a droplet V-The volume of the oil drop q - Charge, in coulombs, carried by the droplet of oil AV - The potential difference across the plates in volts (set it to 500 V) d - The separation of the plates in the condenser (7.6 x 10* m) p - The density of the oil (886 kg/m*) g - Gravitational field strength (9.8 N/kg) p - Barometric pressure (1 atm = 1.01 x 10* pa) n - The viscosity of air (approx. 1.8 x 10* N.s/m?) b - A constant equal to 8.20 x 10* Pa.m w - terminal velocity while falling vr - terminal velocity while risingOil Drop #1 Oil Drop #2 Oil Drop #3 Oil Drop #4 t rise t fall 1 rise t fall t rise t fall rise t fall 4.4 28.13 1.40 6.08 2.79 14.45 2.89 33.44 3.98 38.91 54.53 6.35 2.76 14.36 3.36 38.52 3.64 33.16 35.35 6.83 2.50 16.76 3.80 18.05 2.21 31.4 34.40 7.24 4.90 13.43 1.84 29 62 3.6 47.6 31.0 6.80 4.58 13.95 Average Average 4 39.36 27.36 7.80 2.75 17.78 2.97 37.41 Average Average 26.46 7.50 3.03 12.55 V rise = yit rise,avg V fall = y/t fall,avg 3.65 36.43 Average Average 3.08 14,39 1.68E-04 1.34E-05 V rise = yit rise, avg |V fall = y/t fall.avg 34.36 6.9 3.01 3.95 1.37E-06 1.37E-05 V rise = yit rise,avg | V fall = yit fall.avg 3.11 12.83 1.46E-05 7.25E-05 Average Average 3.25 14.45 V rise = y/t rise, avg | V fall = y/t fall,avg 1.54E-04 3.46E-05 Distance travelled by oil drop: Ay = 0 5e-3_m Plate Voltage: V =_500 v Plate separation: d = 7.6 mm = 7.6e - 3 m Barometric pressure: p = _ 7 60 mm Hg =_ 10105 Pa Resistance (from thermistor): 0 =_ 2.304 x 10# ohms Temperature (from table): T = 20 OC Viscosity (from graph): n = 1. 823 x10.5 N.5/m 1. Calculate the radius of the drop, using equation #4. r = 6+ 209 2. Calculate the mass of the drop, using equation #5. m = pV = p=mr 3. Calculate the electric field strength, using equation #6. E = 4. Calculate the charge of the oil drop, using equation #7.Drop # Charge "q" (10-19 C) (least value for "q") 2 w D V 00 9 10 (greatest value for "q") Gather data from all of the lab groups in your class, to determine the total number of oil drops that have a charge value that falls within the ranges described in the table below: q value Total q value Total q value Total q value Total (10-19 C) Number (10-19 C) Number (10-19 C) Number (10-19 C) Number 0 to 0.25 2.51 to 2.75 5.01 to 5.25 7.51 to 7.75 0.26 to 0.5 2.76 to 3.00 5.26 to 5.50 7.76 to 8.00 0.51 to 0.75 3.01 to 3.25 5.51 to 5.75 8.01 to 8.25 0.76 to 1.00 6 to 3.50 5.76 to 6.00 8.26 to 8.50 1.01 to 1.25 3.51 to 3.75 6.01 to 6.25 8.51 to 8.75 1.26 to 1.50 3.76 to 4.00 6.26 to 6.50 8.76 to 9.00 1.51 to 1.75 4.01 to 4.25 6.51 to 6.75 9.01 to 9.25 1.76 to 2.00 4.26 to 4.50 6.76 to 7.00 9.26 to 9.50 2.01 to 2.25 4.51 to 4.75 7.01 to 7.25 9.51 to 9.75 2.26 to 2.50 4.76 to 5.00 7.26 to 7.50 9.76 to 10.00