Lab Simulation: Find the charge to mass ratio of an electron. 

An electron moves  horizontally with speed 6.00 x 106 m/s between two neutral (uncharged) conducting plates. At t = 1.0 µs, the plates become oppositely  charged, creating a uniform electric field in the region between them. (The time is displayed in units of µs.) The electron deflects vertically upward or downward as it moves to the right a  distance x until it hits a detector. The magnitude and sign of the plate charges, and hence absolute value and direction of electric field, can be changed with the drop down menu below. Let the positive y-direction be up. The y-components of the electric field are shown in the drop down menu below.

N/C (Change electric field)

(Hint: Review Section 21-10 ) 
Question 1
Launch the electron for negative values of the field component. What is the sign of the  charge of the upper plate for these values. The lower plate? 
Question 2
Launch the electron for positive  values of the field component. What is the sign of the charge of the upper plate for these values. The lower plate?  
Question 3
When the field component = +12 N/C, what is the value of the y-component of velocity just before the  electron hits the detector ? What is your answer when  field component = +8 N/C? +4 N/C? (Hint: use the time of flight and Chapter 4
theory of projectile motion.) 
Fill in the data table for a plot of the change in y vs.  Ey. You must measure the change in y using click and drag. You will plot the line by hand using out methods from previous lab, P vs T. That  experiment taught you about using error bars, and finding the max , min and best fit slope using  the special method : at least one/third of the points fall outside area defined by shading  the smallest angle between  the max line segment and the min line segment. To reach this condition, you adjust both the angle and location of the intersection of the above mentioned  two line segments and a third segment   bisecting the angle between between the two outer.   

You will measure   Δy four times for each value of the electric field component.

LC for Δ y = 0.1 m          
-12 -8 -4 4 8 12
 0.32  0.22  0.11  -0.12  -0.23  -0.33
 0.31  0.21  0.10  -0.11  -0.22  -0.34
 0.32  0.22  0.11  -0.11  -0.23  -0.34
 0.31  0.21  0.10  -0.12  -0.22  -0.33
Δybest  = 0.315 Δybest  = 0.215 Δybest   = 0.105 Δybest   = -0.115 Δybest = -0.225 Δybest = -0.335
Δy inst (1.41..)*(LC/10) = 0.01414 = 0.01
Δy inst (1.41..)*(LC/10) =
0.01414 = 0.01
Δy inst ((1.41..)*(LC/10) =
0.01414 = 0.01
Δy inst ((1.41..)*(LC/10) =
0.01414 = 0.01
Δy inst (1.41..)*(LC/10) =
0.01414 = 0.01
Δy inst (1.41..)*(LC/10) =
0.01414 = 0.01
R/N = (0.32 - 0.31)/4

R/N = (0.22 - 0.21)/4
R/N = (0.11- 0.10)/4

R/N = (0.12 - 0.11)/4


R/N = (0.23 - 0.22)/4
R/N = (0.34 - 0.33)/4
error in Δy =  0.01 error in Δy = 0.01 error in Δy =  0.01 error in Δy =  0.01
error in Δy = 0.01
error in Δy = 0.01

Plot Δy vs Ey with vertical errors bars; assume the uncertainty  in Ey is very a small.  Find the error bars using the same method as  in the P vs T, lab--find R/N where in this case the  measurements are “non-simple”-- make the 6 measurements  taking 4 readings each. In each column, take  the difference between the minimum and maximum values  and divide by N (where N =4) . This result is called   R/N.     Compute the average (yBEST) of the 4 measured distances in each case. The  uncertainty  you choose will either be   Δy inst or R/N , which ever is larger.  In either case, the uncertainty is rounded to just one digit and we round the average (“best”) value to the same decimal place. As in the picket fence lab, the result is reported as yBEST ± uncertainty . 


NORMALLY, for  the y measurement, your error Δy inst  in a single   measurement would be L.C./4, where L.C. = smallest division of the grid. Looking at the grid, L.C. = 0.1 m, so the error is 0.025 = 0.03. Thus acceptable values in a  hypothetical experiment  might be 1.40, 1.45, 1.50, 27.20, 27.25 or 27.30, with errors of 0.03 . The 1/100 place can only be filled by a 5 or 0 digit. See the following comment on the error of a  difference that must be used. 


In your
computations, you got the change in y by taking the difference of two values. This raises the accuracy requirements  of our formulas. We are now required to use quadrature. Please go the following page, and scroll  to page 36. Here we get the basic definition  of the LC and error = LC/4. Our LC = 0.1 m; thus our error is 0.025. But because we are taking a difference , we must use quadrature  discussed on page 48 of this page:

Quadrature says the error will be the Pythagorean sum of the two values:

Thus, error = (1.41..)*(LC/4), where 1.41... represents the square root of two. LC = 0.1, producing an error = 0.03536 -
=0.04. Our choices will be different: Suppose you take two values from above and you subtract them to get the final result, how would you report the answer?

2.25 - 27.20 = 0.05. You would report 0.05 plus or minus 0.04.   Apply these concepts to your computation of 
error in y above. 

HOWEVER, there is a different system for the online measurements that modifies the above procedure somewhat. You must change the error. You are not measuring in the way described above because the y position value seems to change continuously by 0.1 as you move the cursor between two adjacent small division marks. Thus,  a more likely uncertainty is  based on an effective least count of LC of 0.1 but with an  error given by 0.1/10 = 0.01. 
Thus,  acceptable values in a  hypothetical experiment  might be 1.41, 1.44, 1.59, 27.23, 27.31 or 17.86 where a result is reported as 1.59 plus or minus 0.01.
From the slope mBEST of the line through the data points, find the charge to mass ratio e/M= -2*mBEST/(change in t)2, where change in t = 0.55 µs.  The error in e/M is thus,  2*Δm/(change in t)2  , where Δm is the error in the slope m. Note: Δm CAN TAKE ON TWO POSSIBLE VALUES,  whichever is larger. The two possible values are (mMAX -mMIN )/2 or LCy/ΔEy, where LCy is the least count (value of the smallest division) on the vertical scale of the graph paper and ΔEy is the value of the base of the triangle  you used to find mBEST . Once you make the correct choice for  Δm, then compare with the theoretical  ratio along the lines of the data sheet below. 

SOLUTION: Let's do some graphing; looking at the table below, we the least count of the graph paper vertical scale is 0.005. This places an interesting discussion of the unlimited resolution you could get if you played your graph pare right. For example, in this case the vertical axis varies in increments of 0.005 across each small division.  Thus, assuming you make a clean small pen mark, your point y point , for Ey = - 10 N ( exact) would lie between , say, 0.260 and 0.265 on the graph paper and for Ey = 8 N, you point point is between -0.215 and - 0.220 depending on your error, you would round off to a certain value on the grid you construct  by dividing the least count be either 2 or 4.  Let's say you take a error  of  LC/2 = 0.0025 by dividing the LC into two halves. 

If the first mentioned y point was below the halfway mark between 0.260 and 0.265, you round  down to 0.260. If the second  mentioned y point was above  the halfway mark between -0.215 and 0.220, you round up  to 0.215. Then you would find the slope as (-0.215 - 0.260)/[ 8 - (-10)] =  

You would have to repeat the same process for  the max and min sloped lines; for more information on how to compute  these two lines' slopes  , here is a special  link ( followed by an email later)  that should help: GO to this page:

and in the left hand column click the button " Lab Manual (All Courses)"
and go to pages 8 - 12, section 1.4. If you read these pages you will
find they are pretty much consistent with what I have been saying
through emails and lab handouts. At the above link, see Figure 2. The
best fit line and maximum and minimum lines have this in common:
approximately equal number of points lie above and below the line in
each case. The minimum and maximum sloped lines have these properties:
"One red line has the largest plausible
slope and one has the smallest."
In the case of the best, minimum and maximum sloped lines, about
the same number of points lie above and below each line. We are computing the uncertainty in the slope using (Mmax - Mmin)/2 in contrast to the recommendation in Figure 3. Note that if the points line up on a line so closely that it's impossible to distinguish  Mmax, Mmin and Mbest, you would use LCy/(change in T) as the slope uncertainty. See discussion of a  previous email with subject line "LAB COMMENT 2 FOR (CHANGE IN Y) VS Ey LAB"

LCy  is the least count of the graph paper in the y direction. LCy  = 0.005

(We decided a the best count was 0.005 m/div, in portrait view. 

ΔEy for mBEST
18 (N)
(see discussion just above table.)
ΔEy for mMAX =
18 (N)
(see discussion just above table.)
for mMIN =
18 (N)
discussion just above table.)
change in Δy for mBEST =
change in Δy for mMAX =


change in Δy 
for mMIN 
mMAX  (-0.195 - 0.245)//[ 8 - (-10)] =  -0.024444.
mMIN  (-0.240 - 0.285)//[ 8 - (-10)] =  -0.02916666.
 (mMAX -mMIN )/2 [-0.024444 - ( -0.02916666 )]/2 = 0.0023613 
LCy/(change in Δy) 0.005/18 = 0.000277777777


  (-0.215 - 0.260)/[ 8 - (-10)] =  -0.026388888.


Choose which ever is larger,  (mMAX -mMIN )/2  or  LCy/(change in Δy); in case,  Δm =0.0023618

e/M -2*mBEST/(change in t)2   =  -2*(0.02638888)/(0.55x10-6)2 = 1.74472x10 11   C/kg
error in e/M 2*Δm/(change in t)2   = 2*(0.0023618 )/(0.55x10-6)2   
0.02638888)*(1.74472x10 11   C/kg ) =

0.1561x10 11 C/kg

Ey (N/C) change in y






Is your result equal to the theoretical value? Hint:  Does the accepted value* (e/M) fall within the range of uncertainty around your best value? In other words, is the following true?
| (e/M)best – e/M | < error in e/M  ?

e/M = 1.75882012 (15) x 10 11 C/kg . 
Go to

|1.74472x10 11 
- 1.75882012 (15)| x 1011  = 0.0140882012 x1011  = 0  < 0.1561x10 11 C/kg
| (e/M)best – e/M | < error in e/M  ?

YES !! Note  we can only compare 
0 and the overall error written as  0.1561x10 11 C/kg  
since the difference
0.0140882012 x1011 = 0 .

*Look up the accepted values of e and M on internet or in the textbook.











Plot the above data in Excel. From the slope of the line through the data points, find the charge to mass ratio e/m. Compare with the theoretical  ratio by finding  the percent error. 

Ey (N/C) change in y