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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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skyluke89

Answer:

[tex]4.8\cdot 10^{-19} C[/tex]

Explanation:

For a drop in equilibrium, the weight is equal to the electric force (in magnitude):

[tex]W = F_e[/tex]

where here we have

[tex]W=1.00\cdot 10^{-14}N[/tex] is the weight of the drop

[tex]F_e[/tex] is the magnitude of the electric force, which can be rewritten as

[tex]F_e = qE[/tex]

where

q is the charge of the oil drop

[tex]E=2.08 \cdot 10^4 N/C[/tex] is the magnitude of the electric field

Substituting into the equation and solving for q, we find the charge of the oil drop:

[tex]q=\frac{W}{F_e}=\frac{1.00\cdot 10^{-14}N}{2.08\cdot 10^4 N/C}=4.8\cdot 10^{-19} C[/tex]

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