Millikan's Oil Drop Experiment graphing problem

[Nitrate]

1. During a Millikan oil drop experiment, a student recorded the weight of 5 different oil drops. A record was also made of the Electric field intensity necessary to hold the drop stationary between thew two horizontal plates.

Data:
Weight x 10^-14 N
1.7
2.9
4.0
5.6
6.1

[Electrostatic field intensity] x 10^5 N/C
1.1
1.8
2.5
3.5
3.8

Q: Charge
n: integer
e: elementary charge
[E]: electrostatic field intensity




2. Slope = Rise/Run
Q = ne



3.
a) Using [E] as the manipulated variable, plot a graph showing the relationship between the weight and the electric field. (I've done this.)

b) Using only the graph, determine the elementary charge (hint: what is the physical meaning of the slope of the graph.)

Work for b) I've calculated the slope using the data [the graph is a straight line]:
Slope = y2 - y1 / x2 - x1
(3.8x10^5 N/C) - (1.1x10^5 N/C) / (6.1 x 10^-14 N) - (1.7x10^-14 N)

and this results in 6.1 x 10^-18 N^2/C [Rounded]

I am now stuck.
I have no clue what the unit N^2/C means, or how the number I found will help me find the elementary charge.

 

[tiny-tim]

Welcome to PF!

Hi Nitrate! Welcome to PF!

(try using the X² icon just above the Reply box )

Work for b) I've calculated the slope using the data [the graph is a straight line]:
Slope = y2 - y1 / x2 - x1
(3.8x10^5 N/C) - (1.1x10^5 N/C) / (6.1 x 10^-14 N) - (1.7x10^-14 N)

and this results in 6.1 x 10^-18 N^2/C [Rounded]

Nooo …

N/C divided by N is 1/C. ​

 

[SammyS]

$$\text{slope}=\frac{3.8\times10^{5} \text{ N/C}-1.1\times10^{5} \text{ N/C}}{6.1\times10^{-14}\text{ N}-1.7\times10^{-14}\text{ N}}=\frac{2.7\times10^{5} \text{ N/C}}{4.4\times10^{-14}\text{ N}}=6.1\bar{3}\bar{6}\times10^{+18}\text{ 1/C}=\frac{1}{\text{?}\text{ C}}$$

You made a little error in your power of ten.