physics

Question

1- An oil droplet with a mass of (8.000x10^-16) kg is moving upward at a constant speed of 2.00 m/s between two horizontal charged plates. If the electric field strength between the plates is (3.50x10^4) N/C, what is the charge on the oil droplet? 

2-During a Millikan oil drop experiment, a student records the weight of five different droplets. A record is also made of the electric field intensity needed to hold each droplet stationary between the horizontal charged plates. 

 

Electric Field Strength (x105N/C)	Weight (x10-14N)
1.1	1.7
3.5	5.6
3.8	6.1
1.8	2.9
2.5	4.0

Graph the recorded data(do on your own, does not need to be submitted but needs to be done in order to answer this question and the next).

Using only the graph, describe how you can determine the elementary charge. 

3- Using your graph and your method described above, the elementary charge is ____x10^-19 C.

Answer 1

(2)

The table gives the electric field (E) and weight (mg)

We know that q = (mg)/E

Electric Fields Strength E (*10^5 N/C)	Weight (mg) (* 10^-14 N)	Charge q = (mg)/E ( * 10^-19 C)
1.1	1.7	1.545
3.5	5.6	1.60
3.8	6.1	1.605
1.8	2.9	1.611
2.5	4.0	1.6

Graph

Plot the table.

Join all the points to a smooth curve.

Using graph, now we deteremine the elementary charge.

Since we can observe from the graph that it is a straight line.

So to find the elementary charge, we find the slope of the line.

Consider the points (3.5,5.6) and (1.1,1.7)

Slope = (5.6 - 1.7)/(3.5 - 1.1)

Slope = 1.60 * 10^-19 C.

Therfore elementary charge is 1.60 * 10^-19 C.

Answer 2

The mass of the oil drop is 8.0 * 10^-16

Electric field of between plates is 3.50 * 10^4 N/c

The electron traveling with constant speed then rate of change of acceleration is zero.

F = Fe - mg

ma = Fe - mg

0 = Fe - mg

Fe = mg

Where Fe is the Electric Field = qE

qE = mg

q = mg/E

q = (8.0 * 10^-16 * 9.8)/(3.50 * 10^4)

q = 2.24 * 10^-19

Therefore the charge of oil drop is 2.24 * 10^-19 C.