Question
In the early 1900s, Robert Millikan used small charged droplets of oil, suspended in an electric field, to make the first quantitative measurements of the electron’s charge. A 0.71-um-diameter droplet of oil, having a charge of , is suspended in midair between two horizontal plates of a parallel-plate capacitor. The upward electric force on the droplet is exactly balanced by the downward force of gravity. The oil has a density of 860 kg/m^3 , and the capacitor plates are 4.5 mm apart.
Part A
What must the potential difference between the plates be to hold the droplet in equilibrium?
Express your answer to two significant figures and include the appropriate units
V=
Explanation / Answer
Given: Diameter of the drop-let of oil d = 0.71m = radious of the droplet = r = 0.355 x 10-6 m volume of enclosed in the drop let of oil V = 4/3 r3 = 4/3 (3.14) (0.355x10-6)3 m3 = 0.187 x10-18 m3 density of the drop let of oil = =860 kg/m3 thus, Mass of th edoplet of oil is
M = V = (860 )(0.187 x10-18 ) = 160.82 x10-18 kg when tis dop -let is suspended in the parallel plates , an amount of electric force F = Eq is acting on this droplet where E is the electric field q is the charge on the electron but , E = V / d where V is the potential difference between the plates d is the distance between the plates : d = 4.5 mm = 4.5 x10-3 m when droplet suspended in the elecric field weight of the droplet should balanced by the elecric force thus, mathematecally ,
M g = E q Mg =Vq / d V = Mg d / q = (160.82 x10-18 )(9.8)( 4.5 x10-3) / (1.6x10-19 ) = 4432.6 x 10-2 volts potentai difference :
V = 44.32 volts potentai difference :
V = 44.32 volts