Millikan's Oil-Drop Experiment Millikan's Qil-Drop Experiment In this lab you will discover the nature of charge in a similar way that Millikan did with his o1l drop experiment. Problem Does there exist a smallest unit of electric charge of which other units are multiples? Equipment charged plates oil droplets potentiometer (to adjust voltage) and power supply s X-ray source New Drop X-ray zap 330V d (between plales) = 0.10m M (of oil droplet) = 1.57x1015 kg g = 9.8 m/s? Procedure 1. Open the Millikan lab. 2. Click the New Drop button to release a droplet. When it gets inside the chamber, click the x-ray button to ionize the chamber and thereby put some charge on the droplet. Millikan's Oil-Drop Experiment 3. Adjust the potentiometer (slider) until the oil drop completely stops moving up or down. 4. Record the voltage in the following table. Voltage (V) Charge (C) ACharge (C) 2 3 4 5 6 7 8 9 - (=] 5. Hit the chamber with another blast of x-rays to change the charge on the oil droplet. Measure the voltage required to stop it again. Repeat this step 8 more times until you have 10 individual values for voltage. If the droplet gets out of the chamber, add a new drop again and continue. Analysis 6. Calculate the amount of charge on the droplet for each voltage. How will you know this? When the droplet completely stops, the electric force up must be exactly equal to the gravitational force down. Fo F, =F, so E,q=mg F, o mg E V _ ) but E = Recall that E, is the Electric d field intensity and q 1s the charge. 56 q= dmg Millikan's Qil-Drop Experiment 8. Finally, calculate the difference between successive charges. 9. Plot a bar graph of the differences against the trial. Conclusion What can you conclude about the existence of a smallest unit of electric charge, of which other charges are multiples? 10. Do some research on the Internet and find an image of the original Millikan apparatus and include it in your write up. 11. Write up your lab and submit it to your teacher. How Did He Do It? How did Millikan find the mass of the droplet? He knew that the oil droplet would accelerate down under the force of gravity until it reached its terminal velocity, the point where the air resistance would stop it from accelerating any more. This of course happens almost instantaneously because the droplet is s0 tiny. The other fact Millikan knew is that the droplet is essentially a sphere shape once it reaches terminal velocity. This is the equation for the terminal velocity of a sphere. 2 . 02222g(D,-D,)r r= n vt is the terminal velocity of the droplet Dy, is the density of the oil or 920 kg/m' DA is the density of the air in the chamber or 1.292 kg/m' n is the dynamic viscosity of the air or 1.80 x 10 * kg/m /s r is the radius of the droplet Using the chamber without charging the droplet, Millikan found the terminal velocity. From that he calculated the radius of the droplet. v, n 0.2222 g (D, -D,) Millikan's Oil-Drop Experiment Using the radius, he calculated the volume. 4 V=nxr 3 Using the volume and the density, he calculated the mass. D=" ; DV m If you are interested, you can use the Pre-Millikan experiment to find the terminal velocity and calculate the mass of each droplet. Does your value match the one you used in the actual Millikan experiment? Figure 1 Robert Millikan fundamental physical constant a measurable natural value that never varies and can be determined by experimentation elementary charge () the magnitude of the electric charge carried by a proton, equal o the absolute value of the electric charge of an electron Fr+rr T Figure 2 A charged oil drop will be suspended between two charged plates when the electric force on the drop balances the gravitational force. The Millikan Oil Drop Experiment Physicists now know that each fundamental particle has a characteristic electric charge that does not change. In fact, the amount of charge is part of the definition of the kind of particle. At the start of the twentieth century, though, physicists still had many questions about the charge of fundamental particles. A big question was, is there a smallest unit of charge that nature will allow, and if so, what is the value of this charge? In this section, you will read about a brilliant experiment aimed at answering this question that resulted in the first measurement of the charge of the electron. Millikan's Experiment Physicist Robert Millikan (Figure 1) set out in 1909 to examine the existence of fun- damental charge using a series of experiments. Millikan's work demonstrated that the electron is a fundamental particle with a unique charge. This electric charge is con- sidered one of a few fundamental physical constantsmeasurable values that can be determined by demonstration and do not varythat define natural laws. Millikan hypothesized that an elementary charge, e, the smallest unit of charge in nature, did exist, and that the charge of the electron equalled this elementary charge. To measure the charge, Millikan used a fine mist of oil droplets sprayed from an atomizer similar to what you may find on perfume bottles. The droplets picked up electric charges due to friction when sprayed from the atomizer. Millikan further hypothesized that the amount of charge any one drop picked up would be a whaole- number multiple of the fundamental charge. To measure the charge on a drop, Millikan used a device called an electrical micro- balance. He allowed the oil drops to fall into a region between two oppositely charged parallel plates. The charges on the plates mean that there is an electric field in the space between the plates, which creates a potential difference between the top plate and the bottom plate. Millikan connected the plates to a series of adjustable batteries sa that he could adjust the magnitude of the electric field and, therefore, the electric force on the droplets. By adjusting the electric force to balance the downward gravi- tational force, Millikan could bring a charged drop of oil to rest in the region between the plates (Figure 2). We can understand how Millikan used the electrical microbalance to determine the charge on an oil drop by comparing the electric and gravitational forces. For a drop of charge g, the electric force from the field is E. = qu If we assume that the charge is positive, then we can charge the plates so that the electric field points upward and gives an upward force to the drop. We can then carefully adjust the potential difference so that the falling drop comes to rest between the plates. When this happens, the electric force balances the gravitational force, and the magnitudes will be equal: Fr=F, tIE - f}ig In Section 7.4, you learned that the electric field between two charged plates depends on the potential difference AV as AV Ad where Ad is the plate separation distance. Therefore, we can solve for the electric charge of the drop as e= _mg q= E B mgAd 17 "4y, where AV}, is the special value of potential difference that balances the drop. If we assume that we can measure the potential difference of our batteries and the plate separation, then we will know the charge of the drop if we can measure the drop's mass. To measure the mass of a drop, Millikan simply switched off the electric field and observed the final speed of the drop as it fell onto the bottom plate. From the final speed, he could calculate the mass of the drop if he accounted for both the gravita- tional force and the force due to air friction. With this information, he could deter- mine the charge on the drop. Millikan repeated his experiment many times, balancing a drop, measuring the voltage, letting the drop fall, and measuring its final speed. When he analyzed the data, he discovered the hypothesized pattern. The values of the charges he measured were whole-number multiples of some smallest value, and no drops had less charge than this value. Millikan concluded that this charge value equalled the elementary chargt: of the electron. In fact, we now think of this positive number as the cha.rge of the proton, but the absolute value is the same for electrons. Later experiments by other researchers confirmed Millikan's results and improved the accuracy of his findings. The current accepted value of the elementary charge to four significant digits is e= 1602 % 107"C With this value, we can connect the charge of an object and the difference in the number of electrons versus protons in the object. If an object has N more protons than electrons, it has a charge 4 given by q = Ne The following Tutorial examines problems involving the elementary charge. AR R 7.6.1 The Millikan Experiment (page 367) In Investigation 7.6.1, you will recreate Millikan's famous experiment using hands-on and online models. Tutorial 1 ' Solving Problems Related to the Elementary Sample Problem 1: Calculating the Charge on an Object Calculate the charge on a small sphere with an excess of Solution: g = Ne 3.2 % 10" electrons. = (3.2 x 10M)(1.602 % 10 " () Given: N = 3.2 = 10" electrons g=-51x10C Required: q Statement: The charge on the sphereis 5.1 % 107C. Analysis: g = Ne; note that in this case e = 1.602 % 10 " C because we are dealing with electrons. Sample Problem 2: The Elementary Charge and an 0il Drop In a Millikan-type experiment, two horizontal plates maintained direction to the electric field. The electric force on the sphere at a potential difference of 360 V are separated by 2.5 cm. must point upward to balance the gravitational force, so the A latex sphere with 8 mass of 1.41 x 107" kg hangs between sphere's charge must be negative. RS e S g i o Wil o {b) Given: r = 2.5cm = 2.5 X 102 m: m = 1.41 X 10" kg; (a) Is the sphere negatively or positively charged? AV, = 360V (b) Calculate the magnitude of the charge on the latex sphere. Required: g (c) Determine the number of excess or deficit particles on the sphere. Analysis: When the sphere is balanced, the electric force ) balances the gravitational force. So, we can use the Solution equation for solving for the electric charge of the oil drop (a) The electric field lines run from positive charges to negative . i L mgAd charges, 5o the field between the plates points downward. a0 M 3 A R R L o AR We know that protons will move in the same direction as Mote that 1V = 1 N-m/Cand 1N = 1 kg' m/s2, an electric field and that electrons will move in the opposite mghd Solution: g = W L (1.41 10 '5kg](9.8 SP)(E.!'} % 10 2m) 360kg-5- 7 g = 9.596 107"9C (two extra digits carried) The actual charge is negative. This value gives the magnitude of the charge. Statement: The magnitude of the charge on the latex sphere 896 x 107" C. Analysis: Use the equation that connects the charge of an object with the number of excess protons in the object: q= Ne. Solution: g = Ne =1 e _ 9.596 x 1078 "~ 1.602 x 10 g N=6 Since the sphere has a negative charge, the excess charges are electrons. {c) Given: g = 9.596 x 10 1 C Statement: There are 6 excess electrons on the sphere. Required: N Practice 1. Calculate the force of repulsion between two plastic spheres placed 110 cm apart. Each sphere has a deficit of 1.2 % 10 electrons. & [ans: 2.7 x 107 N] 2. An oil drop with a mass of 2,48 > 107" kg is balanced between two parallel, horizontal plates 1.7 cm apart, maintained at a potential difference of 260 V. The upper plate is positive. Calculate the charge on the drop in coulombs and as a multiple of the elementary charge. Determine whether there is an excess or a deficit of electrons. =m fans: 1.6 * 107'* ; 108; 10 axcess alectrons] 3. Due to the positive charge of Earth's ionosphere, Earth's surface is surrounded by an electric field similar to the field surrounding a negatively charged sphere. The magnitude of this field is approximately 1.0 > 10 N/C. What charge would an oil drop with a mass of 2.4 x 107" kg need in order to remain suspended by Earth's electric field? Give your answer in both coulombs and as a multiple of the elementary charge. &0 [ans: -2.4 10 " ; 1.5 x 10%] Charge of a Proton 'The elementary charge is the electric charge of a praton, which is equal in magnitude to the absolute value of the electric charge of the electron. Careful experimentation has consistently shown that the two particles have charges that are equal in magnitude. This result is actually a surprise, because the electron and proton have very little else in common, including their masses and the roles they play in the structure of matter. Furthermore, physicists think of the electron as a fundamental particle with no inner workings, but they now view the proton as a combination of more fundamental particles called quarks. The proton consists of three quarks, all of which have charges that are either exactly one-third or two-thirds of the elementary charge. Despite having fractional charges, though, no experiment has detected any combination of quarks in nature that have a total charge whose magnitude is less than e. For this reason, physicists still refer to as the elementary charge. In fact, every subatomic particle that researchers have so far detected has a charge whose magnitude is equal to a whole-number multiple of . Researchers also believe that the amount of charge in an isolated system is conserved like energy. Unlike energy, electric charge does not come in different types and cannot change from one form to another. No interaction can destroy or create electric charge, and the total electric charge of the universe remains constant