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Grade 12Electrostatics

In millikan's experiment, an oil drop of radius 10^-4cm remains suspended between the plates which are 1cm 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 1.5gcm^-3.

Gaurav

8 Years agoGrade 12

4 Answers

Eshan

8 Years ago

Arun

8 Years ago

Dear Gaurav

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

Regards

Arun (askIITians forum expert)

Sami Ullah

8 Years ago

First I will give you the formula to calculate “V”.The droplet is suspended when,

$F_{e}=F_{g}$

$qE=mg$

$q\frac{V}{d}=mg$

$V=\frac{mgd}{q}$

Now all we have to know is the mass of droplet.

$m=\rho \times (Volume)$

$m=\rho \times (\frac{4}{3}\pi\times r^3)$

$m=(1500) \times (\frac{4}{3}\pi\times (10^{-6})^3)$

$m=6.3\times 10^{-15} kg$

Now back to the equation,

$V=\frac{(6.3\times 10^{-15})(10)(10^{-2})}{5\times 1.6\times 10^{-19}}$

$V=787.5$ $Volts$

Please notify if the awnser is correct.The approach is correct but awnser can be different due to calculations.

Further I converted all the measuremests to standard S.I units.

Arun

8 Years ago

Sorry for mistake in calculations

You will get the answer as

$\approx$ 785 Volt

Hope it helps

Regards

Arun (askIITians forum expert)

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