8.0T Fall 001 Electrostatic Force 1 Introduction In this experiment you ll investigate aspects of the electrostatic force. This force has such varied roles as making currents flow in wires, holding atoms and molecules together, making clothes cling, glue stick, etc. One of the aims of this experiment is for you to measure the strength of the Coulomb force. Since r 1 q1q F = rˆ, 4πε 0 r the strength of the Coulomb force in SI units depends on the constant ε 0 called the permittivity of free space, which you can measure in this experiment. You ll set up a horizontal parallel plate capacitor and find the voltage at which a piece of charged aluminum of known weight just lifts off the bottom plate due to the electric force. At that moment, the electric force on the aluminum just balances the force of gravity, r r F electric = F gravity. From the relations between measured or known electrical quantities, material properties and apparatus dimensions you can determine ε 0. Apparatus The parallel plate capacitor is made from two large washers (.5 inch diameter). The washers are separated by 3 pieces of insulating perf-board at about 10 separations. The perfboard is 1.5 mm thick, A small piece of aluminum is placed on the lower washer, and a voltage generated by a high voltage power supply (HVPS) is applied across the washers. Note: the HVPS has a very high internal resistance so you cannot get a serious shock from it. The experiment depends on the smoothness and cleanliness of the washers and the cleanliness of the. This is because charge from the bottom washer must be distributed to the. 1.5 mm Side view Figure 1: perf-board used as spacers on washer
8.0T Fall 001 Electrostatic Force (PD, edited 8/15/01 ND) The should be clean and flat, but we have found that it also needs to be given a rough texture to allow air to pass under it when it lifts off. This is done by using a piece of Kleenex to press the small piece of against a piece of 80 grit sandpaper included with the experiment. Try to handle the with your fingers as little as possible to keep it clean. Tweezers work well. You can see (Figures and 3) that if the isn t flat, either the size of the gap will be reduced or charge will flow out to the part of the nearest the upper washer where larger forces will be exerted than if the were flat. You will use 3 s:.0003 in,.0005 in, and.0007 in. Figure : charge flows to edge of Figure 3: charge flows to center of The lower washer is connected to the negative side of the HVPS, and the upper washer to the positive side. One multimeter set on the +DC 1000 V scale is used to read the HVPS output by connecting the meter leads across the output of the HVPS (the right multimeter in figures 4a and 4b). The second multimeter is connected in series with the gap (the left multimeter in figure 4a and 4b). When the lifts and shorts the gap, the second multimeter will register a voltage and the first multimeter will register a drop in the voltage. You want to determine the voltage across the gap just before the moves. Watch the multimeter readings as you slowly increase the voltage. When the lifts and shorts the gap, the right multimeter reading will drop. Record the voltage just before the drop. washers + - HVPS Figure 4a. Experimental set-up.
8.0T Fall 001 Electrostatic Force 3 Figure 4b: Experimental set-up
8.0T Fall 001 Electrostatic Force 4 (PD, edited 8/15/01 ND) Interpretation of your measurements You want to equate the gravitational and electric forces on the. The gravitational force is just the weight of the. The density ρ of the aluminum material is.7 grams/cc (.7 10 3 kg /m 3 ). The volume of the is then its area (A) times its thickness t. Therefore r F gravity = m total g = ρ ta g Finding the electrical force is a bit subtler. Treat the closely spaced, metal portions of the washers as a parallel-plate capacitor. The electric force on the will be its charge times the electric field it feels. The subtlety lies in calculating that field: The total field in the capacitor is just V/d where V is the applied voltage and d is the spacing between the plates. But only half that field is due to charges on the upper plate; and the other half is due to charges on the lower plate. a E Apply Gauss's Law to a cylindrical Gaussian surface whose upper end is inside the top plate and whose bottom end has area a and is within the gap. (See Figure.) The charge enclosed by the Gaussian surface is a times the charge density σ (charge / area) on the upper plate. So Gauss's Law says that the total electric field in the gap is E = V/d. Now you can find σ by equating the total field, E = V/d, to the field found from Gauss's Law: σ =ε o V d. To find the electric force on the, assume that σ, the charge density on the, is the same as that of the lower washer. Here comes the subtle point. By thinking about the electric field from the charge sheets on the inner surfaces of the plates (positive on top and negative on the bottom) you come to the conclusion that 1/ the total field comes from the upper plate and half from the lower plate. Charges on the feel only horizontal forces from other charges on the bottom plate, so the vertical force on the is due to the electric field of just the top charge sheet: r F electric = Q r E top V εov A = σ A = d d Equating the electrical and gravitational forces, A cancels and you get ε o V A d = ρta g, V = d ρ g t. ε 0
8.0T Fall 001 Electrostatic Force 5 If you plot V vs. t, you should get a straight line whose slope is the coefficient of t. You can calculate the free permittivity of space ε o from your experimental value for the slope: ρ gd ε 0 = slope. Experiment: Data Collection Instructions During the experimental runs, avoid touching the aluminum with your fingers so as to avoid getting any oil on the. Cut a piece of the.0003 in aluminum with scissors. Use the grid on the paper accompanying the to cut a square piece of. Take care not to crumple the edges of the. Take the square piece of aluminum, and press it firmly on the abrasive cloth (located on the experiment board) with a piece of tissue (e.g. Kleenex) between the and your finger. Use the tweezers to transfer the to the bottom washer, and make sure that it s fairly flat. Put the top washer in place and turn up the HVPS all the way. The should lift off and bounce around. Turn down the voltage and repeat. Once the apparatus is working well, carefully and slowly raise the voltage and read the voltmeter when the lifts off. The.0003 in tends to weld to the plates so gentle tapping may help to get the to drop down after you turn down the voltage. Record your result in the data table. Repeat 5 times. Take care that the isn t shorting the gap when you read the meter. Then cut and use the.0005 in and.0007 in s and repeat the above procedure for each case. Use a table like this to enter your experimental data Thickness inch.0003.0005.0007 Thickness meter Voltage Trial #1 Voltage Trial # Voltage Trial #3 Average Voltage Data Analysis: 1. Average the voltages for a given thickness.. You then have 3 numbers: the voltages required to lift.0003 in,.0005 in, and.0007 in thickness. Be sure to calculate the thickness of the in meters. From the previous section, the voltage squared is proportional to the thickness of the where: V = d ρ g t, Thickness of perf-board, d =1.5 10 3 m; 3 3 Density of Aluminum, ρ =.7 10 kg / m ; Acceleration due to gravity, g = 9.8m / s. ε 0
8.0T Fall 001 Electrostatic Force 6 (PD, edited 8/15/01 ND) 3. Plot the voltage squared (V ) vs. the thickness of, t. 4. Your graph of the experimental values of V vs. t should be a straight line with the slope being the factor in parentheses above. Use your data analysis tool to find the slope of the best-fit line. 5. Calculate the free permittivity of space ε 0, from your experimental value for the slope: Parts List: ε o = ρgd slope 1 strip of perfboard 1 plated steel washers 4 #30 tinned copper wire 1 piece of aluminum 1 piece of 80 grit sand paper Figure 5: Experiment parts
8.0T Fall 001 Electrostatic Force 7 Question 1: Conceptual Question A variable high voltage power supply is connected across two metal washers that are separated by a distance d = 1.7 mm. A small piece of aluminum is placed between the washers resting on the lower washer. As the voltage difference across the washers is increased, predict what will happen to a piece of aluminum. Question : Analytical Question a) Using Gauss s Law, find an expression for the electric field for an infinite sheet of charge that is uniformly charged with a positive surface charge density σ > 0. b) Consider two points that are at distances a and b away from the plate with b > a. Explain whether the voltage at the point b is higher or lower than the voltage at the point a. Question 3: Analytical Question a) Using Gauss s Law, find an expression for the electric field between two discs of radius R that are separated by a distance d. The discs have opposite charges that are equal in magnitude placed on them. You may neglect edge effects. Make a sketch of the electric field lines when you include edge effects. b) Suppose a voltage difference V is applied across the two discs. Show that the charge on ε oπr V the positive plate is given by the expression, Q = when you ignore edge d effects. c) Taking into consideration edge effects, the actual expression for the charge is given by ε oπr V Q = f where f is a correction factor that depends on the ration of separation to d plate radius. For circular plates, the factor f depends on d /R as follows: d/r 0. 1.86 0.1 1.167 0.05 1.094 0.0 1.04 0.01 1.03 Explain why the factor f is always greater than 1. Where is the extra charge? f d) Find an expression for the capacitance of the circular discs neglecting edge effects. Now take into consideration the correction factor f. Is the actual capacitance larger or smaller? e) How would you expect your results in parts a) d) to change if you used two washers of inner radius R 1 and outer radius R instead of two discs?
8.0T Fall 001 Electrostatic Force 8 (PD, edited 8/15/01 ND) Question 4: Conceptual Question Why is the approximation using Gauss s law reasonable for the experiment? Look at the electric field simulation and estimate the error you can expect by introducing this approximation. Question 5: Analytical Question: Find an expression for the capacitance of two parallel discs of radius R that are separated by a distance d. You may neglect edge effects. What is the total stored energy when a voltage difference V is applied across the discs? Question 6: Analytical Question Suppose a voltage difference V is applied across two washers that are separated by a distance d. A small piece of aluminum of area A is placed between the two washers resting on the lower washer. a) Find an expression for the charge per unit area σ on the aluminum. b) Should the equation for the electric force on the aluminum be F σ A =? Explain your reasoning. ε o F σ A = or ε o Question 7: Analytical Question Suppose a voltage difference V is applied across two washers that are separated by a distance d. A small piece of aluminum is folded n times. Each fold has area A and thickness t. The folded aluminum is placed between the two washers resting on the lower washer. The density of aluminum is ρ. The acceleration due to gravity is g. a) Find an expression for the gravitational force on the n-folds of aluminum. b) If the upper washer is at a higher voltage than the lower washer, what is the direction of the electric force on the aluminum? c) Predict what will happen to the aluminum as the voltage difference between the two washers is increased. d) When the aluminum just starts to move, use Newton s Second Law to find a relation between the voltage difference V and the number of folds n of the aluminum. e) Explain why your result in part d) does or does not depend on the area of the aluminum. f) Suppose that 3 folds of aluminum just start to move when the voltage difference is V = 500V. Let the distance d =1.7 10 3 m and the thickness t = 7 10 6 m. The density 3 3 of aluminum is ρ =.7 10 kg / m. The acceleration due to gravity is g = 9.8m / s. What value do you get for the electric permittivity of free space ε o?
8.0T Fall 001 Electrostatic Force 9 Question 8: Conceptual Question Think about this question while you conduct the experiment: Why does the voltage reading of the multimeter that is connected across the high voltage power supply (HVPS) drop when the jumps up? (Hint: Observe that when the jumps, it never leaves the bottom washer.) Question 9: Analytical Question In this experiment, you measured the voltage difference across the washers when the aluminum just started to jump up. When the jumps up, it connects the two plates, short-circuiting the capacitor. Just when the current starts to flow, the voltage difference across the HVPS drops from 340V to 300V. The HVPS has an internal resistance r int = 3.1 10 6 Ω. When the jumps, current now flows through the second multimeter as well as the first. The second multimeter registers a value of 300V as well. The multimeters, when set on the +DC 1000 V scale, have an internal resistance, R = 0.0 10 6 Ω. a) Just before the aluminum jumps, calculate the current that flows through the multimeter connected to the output of the HVPS. b) What is the electromotive force supplied by the HVPS? c) Draw a circuit diagram for the experimental apparatus just after the aluminum jumps. d) Calculate the total current that flows from the HVPS after the aluminum short circuits the capacitor (both multimeters show 300 V). e) Did the electromotive force of the HVPS change just after the aluminum short circuits the capacitor? Justify your answer by calculating the electromotive force. [???] Question 10: Data Analysis Question Can you use zero volts as one of the data points on your graph of ( V ) vs. n. Explain why or why not. Question 11: Data Analysis Question a) Based on your experimental data, describe which measurements contributed significantly to the total error in your data. b) What is your estimate for the fractional error associated to each of these quantities? c) What is the total fractional error for the permittivity of free space ε o?