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Conceptual Physics | 12th Edition | ISBN: 9780321909107 | Authors: Paul G. Hewitt
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In 1909, Robert Millikan was the first to find the charge

Chapter 14, Problem 10P

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QUESTION:

In 1909, Robert Millikan was the first to find the charge of an electron in his now-famous oil-drop experiment. In that experiment tiny oil drops were sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops were observed with a magnifying eyepiece, and the electric field was adjusted so that the upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended, upward force ?qE? just equaled ?mg?. Millikan accurately measured the charges on many oil drops and found the values to be whole-number multiples of 1.6 × 10?19C—the charge of the electron. For this he won the Nobel Prize. a. If a drop of mass 1.1 × 10?14 kg remains stationary in an electric field of 1.68 × 105 N/C, what is the charge of this drop? b. How many extra electrons are on this particular oil drop (given the presently known charge of the electron)?

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QUESTION:

In 1909, Robert Millikan was the first to find the charge of an electron in his now-famous oil-drop experiment. In that experiment tiny oil drops were sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops were observed with a magnifying eyepiece, and the electric field was adjusted so that the upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended, upward force ?qE? just equaled ?mg?. Millikan accurately measured the charges on many oil drops and found the values to be whole-number multiples of 1.6 × 10?19C—the charge of the electron. For this he won the Nobel Prize. a. If a drop of mass 1.1 × 10?14 kg remains stationary in an electric field of 1.68 × 105 N/C, what is the charge of this drop? b. How many extra electrons are on this particular oil drop (given the presently known charge of the electron)?

ANSWER:

Step 1 of 3 a)we know that F E= q Millikan's experiment was meant to have the drops fall at a constant rate. At this constant rate, the force of gravity on the drop and the force of the electric field on the drop are equal: F = F up down F = qE and F = mg up down Q is the charge of an electron, E s the electric field, m is mass of the droplet, and g is gravity. Eq=mg

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