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MI W. Charged Parallel Plates and Force puelu We have already studied the electric field line pattern between two charged parallel plates. The electric field lines point away from the positive plate towards the negative plate. There is a slight bulge at each edge of the plates. postttve plate +51 1F! b negative plate In the region between the plates, the electric field lines are parallel and evenly spaced. This indicates that the electric field has the same magnitude and direction at all points. If a positive point charge is placed between the plates. the net force on the charge will be in the same direction as the electric field lines. A negative point charge will move in a direction opposite the electric field line direction. The amount of the electric force on the charged particle can be determined by F8 = qE. If the electric force is the only force acting on the charged object, then the acceleration of the object can be determined from Newton's second law: F = ma. For example, if the electric field strength between the charged plates is 10.0 WC, and the object has a charge of q =+ 2.00 C and a mass of m => 4.00 x104 kg. then the acceleration of the particle is a z = E =EZ'UOC110'0NIC): 500 X104 111.152. The direction of the acceleration is in the same direction as the force. For m m 4.00x104kg the diagram above, the charge would accelerate in the down direction. If the charged plates are oriented in an up and down direction as shown above, then the particle also experiences a force of gravity Fg acting downwards. The magnitude of this force is Fg = mg = 3.92 x10'3 N. Since both the electrical force and the force of gravity are acting downwards. the net force is Fnet = Fe + F9 = qE + mg = (2.00C)(10.0N/C) + (4.00x10'4)(9.8N/kg) = 20.0 N The effect of the gravitational force is negligible compared to the electric force. IMAGE 48 by 48 pixels Charged Parallel Plates and Kinematics In our study of kinematics, we applied three equations to the motion of objects. average velocity = displacement time At change in velocity average accleratin V2 - V1 _ Lv elapsed time average velocity = sum of velocities 1 and 2 2 In the previous example, we saw that a charge of +2.00 C accelerated downwards at a rate of 5.00 x 104 m/s downwards, or -5.00 x 104 m/s2. If the particle accelerates downwards for 6.00 us (6.00 x 10- s), the the change in velocity is AV= aAt= (-5.00x104 m/s2)(6.00x10-6) = -0.300 m/s If the initial velocity of the charged object is zero, then after 6.00 ms, the new velocity of the particle is 0.300 m/s and the average velocity is Vave = _(0.00m/s + -0.300m/s) = -0.150 m/s. The downward displacement of the charged particle is AX = VaveAt = (-0.150m/s)(6.00x10-6) = = -9.00 x 10-7 m.MI mm The Millikan Experiment mam At the turn of the century, when the understanding of electric forces was beginning to increase, two fundamental questions arose regarding the nature of electric charge. '1. Does there exist in nature a smallest unit of electric charge of which all other units are multiples? 2. If so, what is this elementary charge, and what is its magnitude, in coulombs? To answer these questions, an American physicist, Robert Andrews Millikan (18681953) devised and performed a series of very creative experiments. He reasoned that the elementary charge would be the charge on an individual electron. He assumed, further, that when tiny oil droplets are sprayed in a fine mist from an atomizer, they become electrically charged by friction, some acquiring an excess of a few electrons, others acquiring a deficit. Although there was no way of knowing how many extra electrons there were on any oil drop, or how many were missing, Millikan hypothesized that if he were able to measure the total charge on any drop, it would have to be some integral multiple of the elementary charge. To measure this charge, Millikan made use of the uniform electric field in the region between two charged parallel plates. He charged the plates by connecting each to opposite terminals of a large bank of storage batteries. There was a small hole in the upper plate and a mist of oil droplets passed through this small hole. By adjusting the voltage applied to the plates, it was possible to change the electric field strength and thus the electrical force acting on the droplet. By making this adjustment, it was possible to make the strength of the electric force equal and opposite to the strength of the gravitational force. Thus the droplet was suspended. If the droplet had a positive charge, then it would be suspended between the charged plates as shown below. negative plate Fe When the oil droplet is in balance, then FB = F9, and qE = mg. The charge on the charged particle can then be determined W by _E. F, q E pa s1Live plate It was possible for Millikan to determine the electric field strength by knowing the voltage applied to the plates and the distance between the plates. When the electric field is turned off, the charged particle begins to fall under the influence of gravity. It soon reaches terminal velocity because of air resistance. The theory of air resistance tells us that terminal velocity depends on the radius of the particle. Once the radius was determined then the mass of the particle could be calculated from the known density. Thus Millikan was able to measure the charge of the droplet. 2:\": Millikan and the Elementary Charge Millikan repeated his experiment many times over. He generated a long list of values for the total electric charge on the droplets. The list showed a very significant pattern. All of the values were multiples of some smallest value. Some of the droplets had this smallest value on them. but none had less. Millikan assumed that this smallest value represented the smallest quantity of electric charge possible, the charge on an electron, or the elementary charge. Accurate measurements of the elementary charge have yielded values very close to that determined by Millikan. The currently accepted value for the elementary charge is 1.602 177 33 x 10' 19 C. The value we will use in our calculations is 1.60 x 10'19 C Thus the total charge, (7, on any object is a whole number multiple, N, of this elementary charge. e. q = Ne If the total charge on an object is 1C, then it is possible to determine the number of elementary charges in this 10. l N: =6.25310182 160x10'l9CIe IMAGE 48 by 48 Summary pixels For a charged particle between two charged plates, the net force on the particle is the sum of the electrical force and the gravitational force on the particle: Fnet = Fe + Fg The electrical force is Fe = qE and the gravitational force is F = mg. The acceleration of the particle is found using Newton's second law: Fnet = ma. We can apply the equations, as discussed in kinematics, for the motion of an charged object in an electric field. average velocity = displacement Vay = 42 - X] - 4x time change in velocity average accleratin day = 12 - VI elapsed time average velocity = sum of velocities 1 and 2 2 Wave = = ( 1 + 12 ) In his experiment, Millikan attempted to determine the answer to two questions. 1. Does there exist in nature a smallest unit of electric charge of which all other units are multiples? 2. If so, what is this elementary charge, and what is its magnitude, in coulombs? Using a charged oil droplet between two charged plates, Millikan was able to suspend the particle between the two plates. This meant that the sum of the upward electrical force equaled the downward pull of gravity. Fe = Fa, and qE = mg. The charge on the charged particle can then be determined by q = mg E The currently accepted value for the elementary charge is 1.602 177 33 x 10-19 C. Thus the total charge, q, on any object is a whole number multiple, N, of this elementary charge, e. q = Ne The number of elementary charges in this 1C is N =- = 6.25 x 1018 2 1.60x10 "CleThe physics of the Miiiikah experiment and forces In a Millikan type experiment, two horizontal plates are separated by a distance of 2.5 cm, and the upper plate is positive. The voltage to the plates is adjusted so that a 1.5 x 10'15 kg sphere remains stationary. 3. Draw a freebody diagram showing the forces acting on the charged sphere q. Label the force vectors with appropriate symbols identifying the forces involved. b. Is the sphere charged negatively or positively? c. If the electric field intensity is 1.84 x 104 NiC, what must be the charge on the sphere? d. How many elementary charges does this correspond to? e. If the electric field is turned off, and the charged sphere begins to fall, what will its velocity be after a time of 1.0 ms? Assume that air resistance over this short time is negligible