Millikan measured the electron's charge by observing tiny charged oil drops in an electric field. Each drop had a charge imbalance of only a few electrons. The strength of the electric field was adjusted so that the electric and gravitational forces on a drop would balance and the drop would be suspended in air. In this way the charge on the drop could be calculated. The charge was always found to be a small multiple of 1.60 × 10-19 C. Find the charge on an oil drop weighing 1.00 × 10-14 N and suspended in a downward field of magnitude 2.08 × 104 N/C.
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Answer:
Explanation:
For a drop in equilibrium, the weight is equal to the electric force (in magnitude):
where here we have
is the weight of the drop
is the magnitude of the electric force, which can be rewritten as
where
q is the charge of the oil drop
is the magnitude of the electric field
Substituting into the equation and solving for q, we find the charge of the oil drop:
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Explanation :
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Answer:
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Explanation:
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as 0.700c and its length as 78.0 meters, then engineers aboard
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