- 2 Given That A Any V 2gp Show That For An Oil Drop With Mass M Which Has Reached Constant Or Terminal Velocity 1 (1.04 MiB) Viewed 54 times
- 2 Given That A Any V 2gp Show That For An Oil Drop With Mass M Which Has Reached Constant Or Terminal Velocity 2 (269.69 KiB) Viewed 54 times
- 2 Given That A Any V 2gp Show That For An Oil Drop With Mass M Which Has Reached Constant Or Terminal Velocity 3 (214.6 KiB) Viewed 54 times
For an oil drop with mass, m, which has reached constant or terminal velocity, Vs. the upward retarding force equals the downward gravitational force and F = mg - 6 (2) Now let an electrical field, E, be applied between the plates in such a direction as to make the drop move upward with a constant velocity, Ve. The viscous force again opposes its motion but acts downward in this case. If the oil drop has an electrical charge, q, when it reaches constant velocity the forces acting on the drop are again in equilibrium and Eq = mg - 6an, Solving Equation (3) for mg and equating this to Equation (2) we obtain q- (v +v) The electric field, E, is obtained by applying a voltage, V, to the parallel plates of the condenser which are separated by a distance, d. Therefore, 6 and q= v,+V) (5) From Equation (5) it is seen that for the same drop and with a constant applied voltage, a change in q results only in a change in ve and ON AV 6 and (6) When many values of Ave are obtained, it is found that they are always integral multiples of a certain small value. Since this is true for Ave, the same must be true for Aq; that is, the charge gained or lost is the exact multiple of a unit charge. Thus the discreteness of the electric charge may be demonstrated without actually obtaining a numerical value of the charge.
In Equation (5) all quantities are known or measurable except a, the radius of the drop. To obtain the value of a, Stokes' law is used. It states that when a small sphere falls freely through a viscous medium it acquires a constant velocity, 2ga (P-P) 971 (7) where p is the density of the oil, pı, the density of the air, and ŋ as stated previously, is the viscosity of the air. Since the density of the air is very much smaller than the density of the oil, pı is negligible and Equation (7) can be solved to give 9V 123p (8) Substituting this value of a in Equation (5) gives an expression for q in which all the quantities are known or measurable: 60 97 9- 9- V2gp v. +v. : (9) Using the MKS system throughout the charge q will be in coulombs when a = radius of the sphere in meters n= viscosity of air in Newton-sec./m2 (see Fig. 3) p=973 Kg/m3 = density of silicone oil g = acceleration of gravity = 9.8 m/sec2 VE = velocity in meters/sec d = distance in meters = 5.0 x 10-2m V = potential difference between plates in volts q=charge in coulombs