Experiment 4: Electrostatic Force

Transcription

1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8. Sprin 3 OBJECTIVE Experiment 4: Electrostatic Force To measure ε, the permittivity of free space. INTRODUCTION Electrostatic force plays a variety of roles in Nature. Examples include makin currents flow in wires, holdin atoms and molecules toether, makin clothes clin, lue stick, etc. The strenth of the electrostatic force is proportional to the product of a chare (measured in coulombs) and the electric field (measured in volts/meter) produced by the other chares. In this experiment, you ll set up a horizontal parallel plate capacitor and find the voltae at which a piece of aluminum foil of known weiht just lifts off the bottom plate. The electric force then just balances the force of ravity: F = F e (4.1) In addition to a sinle thickness of foil, you will also be usin folded foils which increases the thickness by n folds. The electric field between the plates is proportional to the surface chare density on the plates. When the electric field is measured in SI units, the constant of proportionality between the field and the surface chare density is determined by ε, the permittivity of free space. From the relations between measured or known electrical quantities, material properties and apparatus dimensions you can determine the constant ε. PROCEDURE To carry out the experiment, you will need the parallel plates that are made from two lare washers (.5 inch diameter). The washers are separated by 3 pieces of insulatin perf-board havin a thickness of 1.5 mm, at about 1 separations, as illustrated in Fiure 4.1: E4-1

2 Fiure 4.1 Perf-board used as spacers on washer 1. Place a small piece of aluminum foil on the lower washer, and apply a voltae enerated by HVPS (hih voltae power supply) across the washers. Note: (1) The HVPS has a very hih internal resistance so you cannot et a serious shock from it. The experiment depends on the smoothness and cleanliness of the washers and the cleanliness of the foil. This is because chare from the bottom washer must be distributed to the foil. () The foil should be clean and flat, but it also needs to be iven a rouh texture to allow air to pass under it when it lifts off. This is done by usin a piece of Kleenex to press the small piece of foil aainst a piece of 8 rit sandpaper included with the experiment. Try to handle the foil with your finers as little as possible to keep it clean. Tweezers work well. You can see (Fiures 4. and 4.3) that if the foil isn t flat, either the size of the ap will be reduced or chare will flow out to the part of the foil nearest the upper washer where larer forces will be exerted than if the foil were flat. You will use 3 foils:.3 in,.5 in, and.7 in. Fiure 4. Chare flows to ede of foil Fiure 4.3 Chare flows to center of foil. Connect the lower washer to the neative side of the HVPS, and the upper washer to the positive side. One multimeter set on the +DC 1 V scale is used to read the HVPS output by connectin the meter leads across the output of the HVPS (the riht multimeter in fiures 4.4a and 4.4b). The second multimeter is connected in series with the ap (the left multimeter in fiure 4.4a and 4.4b). E4-

3 washers + - HVPS Fiure 4.4a Experimental setup 3. When the foil lifts and shorts the ap, the second multimeter will reister a voltae and the first multimeter will reister a drop in the voltae. You want to determine the voltae across the ap just before the foil moves. Watch the multimeter readins as you slowly increase the voltae. When the foil lifts and shorts the ap, the riht multimeter readin will drop. Record the voltae just before the drop. Fiure 4.4b Experimental setup E4-3

4 ANALYSIS You want to equate the ravitational and electric forces on the foil. The ravitational force is just the weiht of the foil. The density of the aluminum foil material is ρ = rams/cc (.7 1 k/m ). The volume of the foil is then its area (A) times its thickness t. Therefore the ravitational force is F = F = m = ρ ta (4.) Findin the electrical force is a bit subtler. Treat the closely spaced, metal portions of the washers as two equal and opposite parallel disks. The electric force on the foil will be its chare times the electric field it feels. The subtlety lies in calculatin that field. The total field in the capacitor is just E = V d where V is the applied voltae and d is the spacin between the plates. But only half that field is due to chares on the upper plate; the other half is due to chares on the lower plate. In II. Coulomb s Law -Worked Examples, we have shown that the electric field between two infinite oppositely chared parallel plates is uniform and iven by E = σ (4.3) ε o where σ is the chare density (chare/area) on the upper plate. Since the field between the plates is approximated as uniform, the electric potential difference between the plates (the voltae difference) is The chare density σ is therefore V = Ed (4.4) V σ = ε o (4.5) d To find the electric force on the foil, assume that σ, the chare density on the foil, is the same as that of the lower washer. Here comes the subtle point. By thinkin about the electric field from the chare sheets on the inner surfaces of the plates (positive on top and neative on the bottom) you come to the conclusion that half the total field comes from the upper plate and half from the lower plate. Chares on the foil feel only horizontal forces from other chares on the bottom plate, so the vertical force on the foil is due to the electric field of just the top chare sheet: V εo V A Fe = Fe = QfoilEtop = σ A = (4.6) d d E4-4

5 Equatin the electrical and ravitational forces: The area A cancels and we obtain εo d V A = ρ ta (4.7) d ρ (4.8) ε V = t If you plot V vs.t, you should et a straiht 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: d ρ ε = (4.9) slope E4-5