F.6 phy. electrostatics 2

chanel

2008-12-07 6:38 am

Millikan produced an oil drop of radius 1.0X10^-6 m and introduced it between two horizontal plated separated by a distance of 5mm. The upper plate was at a potential V higher than the lower plate. It was found that by adjusting V to a value of 385V, the oil drop would remain stationary under the combined influence of the gravitational and electric forces. Find the charge on the oil drop(include the sign).

The density of the oil is 900kgm^-3.

回答 (1)

Audrey Hepburn

2008-12-07 6:46 am

✔ 最佳答案

Assuming the oil drop is a sphere.

Mass of the sphere, m

= Density X volume

= 900 X 4/3 π(1.0 X 10-6)3

= 3.77 X 10-15 kg

Now, gravitational force is acting downward, while the electric force is acting upward

So, the two forces balance each other.

We have,

qE = mg

qV/d = mg

q = mgd / V

q = (3.77 X 10-15)(10)(5 X 10-3) / 385

Charge on the oil drop, q = 4.90 X 10-19 C

Since the upper plate is at a higher potential. And the electric force on the oil drop is upward.

So, the oil drop is negatively charged.

Therefore, the charge of the oil drop is -4.90 X 10-19 C

參考: Myself~~~

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