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An oil drop i...

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An oil drop is found floating freely between the plates of a parallel plate capacitor the plates being horizontal and the lower plate carrying charge of +Q. The area of each plate is A and the distance of separation between them is D. The mass of oil drop is M . In CGS unit what must be the charge on the drop

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Q. A charged oil drop od mass

2.5 \times 10^{- 7} k g

is in space between the two plates, each of area

2 \times 10^{- 2} m^{2}

of a parallel plate capacitor. When the upper plate has a charge of

5 \times 10^{- 7} C

and the lower plate has an equal negative charge then the oil remains stationery. The charge of the oil drop is (take,

g = 10 {m / s}^{2}

)

Q. Two horizontal plates charged with

+ q

- q

charges are having an area of

A

. A charged drop of oil is suspended in equilibrium position between the plates, then the charge on the oil drop will be :

Q. In a Millikans oil drop experiment, an oil drop of mass

0.64 \times 10^{- 14} k g

, carrying a charge

1.6 \times 10^{- 19} C

remains stationary between two plates separated by a distance of 5 mm. Given

g = 9.8 m / s^{2}

; the voltage that must be applied between the plates being

Q. In Millikan’s oil drop experiment, an oil drop of radius

r

Q

is held in equilibrium between the plates of a charged parallel plate capacitor when the potential difference is

V

. To keep a drop of radius

2 r

2 Q

in equilibrium between the plates ,the potential difference required will be

Q. In a Millikan's oil drop experiment , an oil drop of mass

0.64 \times 10^{- 14} kg

carrying a charge

1.6 \times 10^{- 19} C

remains stationary between two plates separated by a distance of

5 mm .

The voltage that must be applied between the plates is-
[Take

g = 9.8 {ms}^{- 2}

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