In Millikan\'s experiment, a drop of oil (density ? = 0.925 g/cm3) carries an ex

Question

In Millikan's experiment, a drop of oil (density ? = 0.925 g/cm3) carries an excess charge of 8 electrons. The magnitude of the electric field needed to levitate the oil drop is 2.78 x 106 N/C. What is the radius of the oil drop (assumed to be spherical)?

Express the result in the unit [µm] and to three significant figures. If you must use scientific notation, please enter as follows: e.g. 0.000123 = 1.23E-4. Only answer in numerical values, no units. For negative numbers, leave no space between the negative sign and the number: e.g. right: -1.00, wrong: - 1.00.

Explanation / Answer

you know the potential energy of the field from qE
this is (8*the charge of an electron)*E

8 * 1.602 x 10-19 * 2.78 x 106 = 3.56 x 10 -12

this equals mg, since the force of the electric field must equal the force of gravity since its levitating

3.56 x 10-12 / 9.81 = m = 3.63 x 10-13 g

now solve for the volume of the oil drop

v = m/density = 3.926 x 10-13 cm3

its spherical so the formula for volume is 4/3 r3 solve for r

r = 4.542x10-5m = 45.42 m

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