Question

In Millikan's experiment, an oil drop of radius 10^(-4) cm remains suspended between the plates which are 1 cm apart. If the drop has charge of 5e over it, calculate the potential difference between the plates. The density of oil may be taken as 1.5 gcm^(-3).

Answer 1

If the oil drop is suspended stationary, then there would not be any viscous or drag force. If we draw the free body diagram for the oil drop, then the electric force acts in upward direction and the weight of the drop acts in downward direction. Which have to be equal. Therefore,

Fup =Fdown

qV/d = mg

5*1.6*10^-19 * V/10^-2 = (4/3 * pi * R^3)* rho * 10

put R = 10^-6 m

rho = 1500 kg/m^3

then you will find V = 7.86 * 10^-36 Volt

Answer 2

Answer:

769.60

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