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whatis oil drop experiment? what is its use?who proposed that?

whatis oil drop experiment? what is its use?who proposed that?

Grade:11

2 Answers

saiteja007
14 Points
5 years ago
The oil drop experiment was an experiment performed by Robert A. Millikan and Harvey Fletcher in 1909 to measure the elementary electric charge (the charge of the electron). The experiment entailed observing tiny charged droplets of oil between two horizontal metal electrodes.
saiteja007
14 Points
5 years ago

Initially the oil drops are allowed to fall between the plates with the electric field turned off. They very quickly reach a terminal velocity because of friction with the air in the chamber. The field is then turned on and, if it is large enough, some of the drops (the charged ones) will start to rise. (This is because the upwards electric force FE is greater for them than the downwards gravitational force Fg, in the same way bits of paper can be picked by a charged rubber rod). A likely looking drop is selected and kept in the middle of the field of view by alternately switching off the voltage until all the other drops have fallen. The experiment is then continued with this one drop.

The drop is allowed to fall and its terminal velocity v1 in the absence of an electric field is calculated. The drag force acting on the drop can then be worked out using Stokes' law:

F_{d} = 6\pi r \eta v_1  \,

where v1 is the terminal velocity (i.e. velocity in the absence of an electric field) of the falling drop, η is the viscosity of the air, and r is the radius of the drop.

The weight w is the volume D multiplied by the density ρ and the acceleration due to gravity g. However, what is needed is the apparent weight. The apparent weight in air is the true weight minus the upthrust (which equals the weight of air displaced by the oil drop). For a perfectly spherical droplet the apparent weight can be written as:

\boldsymbol{w}=\frac{4\pi}{3}r^3(\rho-\rho_{air})\boldsymbol{g}

At terminal velocity the oil drop is not accelerating. Therefore the total force acting on it must be zero and the two forces F and w must cancel one another out (that is, F = w). This implies

r^2 = \frac{9 \eta v_1}{2 g (\rho - \rho _{air})}. \,

Once r is calculated, w can easily be worked out.

Now the field is turned back on, and the electric force on the drop is

F_E = q E \,

where q is the charge on the oil drop and E is the electric field between the plates. For parallel plates

E = \frac{V}{d} \,

where V is the potential difference and d is the distance between the plates.

One conceivable way to work out q would be to adjust V until the oil drop remained steady. Then we could equate FE with w. Also, determining FE proves difficult because the mass of the oil drop is difficult to determine without reverting to the use of Stokes' Law. A more practical approach is to turn V up slightly so that the oil drop rises with a new terminal velocity v2. Then

q\boldsymbol{E}-\boldsymbol{w}=6\pi\eta\boldsymbol{(r\cdot v _2)}=\left|\boldsymbol{\frac{v_2}{v_1}}\right|\boldsymbol{w}.

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