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 0.79-μm-diameter droplet of oil, having a charge of +e, 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/m3, and the capacitor plates are 6.0 mm apart.
What must the potential difference between the plates be to hold the droplet in equilibrium?
electric force Fe = E*q = (V/d)*q
V = potential difference
d = seperation between plates
gravitational force Fg = -m*g
m = mass of drop = D*V
D = density
V = volume = (4/3)*pi*r^3
r = radius of drop = 0.79/2 = 0.395*10^-6 m
IN equilibrium Fnet = 0
Fe + Fg = 0
(V/d)*q - D*(4/3)*pi*r^3*g = 0
V = D*(4/3)*pi*r^3*g*d/q
V = 860*(4/3)*pi*(0.395*10^-6)^3*9.8*6*10^-3/(1.6*10^-19)
potential difference V = 82 Volts
DONE please check the answer. any doubts feel free to ask
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