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2 years ago

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In Milikan’s experiment, a drop of radius of 1.64 μm and density 0.851 g/cm3 is suspended in the lower chamber when a downward-p

ointing electric field of 1.92 * 105 N/C is applied.
What is the weight of the drop?

Find the charge on the drop, in terms of e.

How many excess or deficit electrons does it have?

1 answer:

jek_recluse [69]2 years ago

3 0

(a) The weight of the drop is 1.54 x 10⁻²⁵ N,

(b) The the charge on the drop, in terms of e is 5 x 10⁻¹²e and

(c) The excess electrons is 5 x 10⁻¹² electron.

<h3>Weight of the drop</h3>

The weight of the drop is calculated as follows;

Volume of the drop; V = ⁴/₃πr³

V = ⁴/₃π(1.64 x 10⁻⁶)³ = 1.845 x 10⁻¹⁷ m³

mass = density x volume

mass = 0.851 g/cm³ x 1.845 x 10⁻²³ cm³ = 1.572 x 10⁻²³ g = 1.57 x 10⁻²⁶ kg.

Weight = 1.57 x 10⁻²⁶ kg x 9.8 m/s² = 1.54 x 10⁻²⁵ N.

<h3>Charge on the drop</h3>

q = F/E

q = (1.54 x 10⁻²⁵ N)/(1.92 x 10⁵)

q = 8.01 x 10⁻³¹ C

1.6 x 10⁻¹⁹ C = 1e

8.01 x 10⁻³¹ C = ?

= 5 x 10⁻¹²e

<h3>Excess electron on the drop</h3>

1.6 x 10⁻¹⁹ C ------- 1 electron

8.01 x 10⁻³¹ C ------- ?

= 5 x 10⁻¹² electron

Thus, the weight of the drop is 1.54 x 10⁻²⁵ N, the the charge on the drop, in terms of e is 5 x 10⁻¹²e and the excess electrons is 5 x 10⁻¹² electron.

Learn more about charge of electron here: brainly.com/question/9317875

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If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2

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