(Hint: Review Section 2110 )
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 ycomponent 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.)
Question 4: MEASURING e/M = ELECTRON CHARGE TO MASS RATIO:
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 
Δy_{best }= 0.315  Δy_{best }= 0.215  Δy_{best }= 0.105  Δy_{best } = 0.115  Δy_{best }= 0.225  Δy_{best }= 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 =0.0025 
R/N = (0.22  0.21)/4 =0.0025 
R/N = (0.11 0.10)/4 =0.0025 
R/N = (0.12  0.11)/4 =0.0025

R/N = (0.23  0.22)/4 =0.0025 
R/N = (0.34  0.33)/4 =0.0025 
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 E_{y} with vertical errors bars; assume the uncertainty in E_{y}
is very a small. Find the error bars using the same method as in the
P vs T, labfind R/N where in this case the measurements are
“nonsimple” 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 (y_{BEST}) 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 y_{BEST} ± uncertainty .
ERRORS FOR Δ y: PLEASE READ CAREFULLY, SINCE THE ERROR IS
NOT WHAT IT SEEMS; it is 0.01 m.
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.
ERROR OF THE DIFFERENCE
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 http://www.physics.upenn.edu/~uglabs/,
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:
http://www.physics.upenn.edu/~uglabs/
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
m_{BEST }of the line through the data
points, find the charge to mass ratio e/M= 2*m_{BEST}/(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 (m_{MAX} m_{MIN
})/2 or LC_{y}/ΔE_{y}, where LC_{y}
is the least count (value of the smallest division) on the vertical scale of the
graph paper and ΔE_{y} is the value of the base of the
triangle you used to find m_{BEST} . 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)] =
0.025.
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:
http://www.physics.upenn.edu/~uglabs/
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"
LC_{y} is the least count of the graph paper in the y
direction. LC_{y} = 0.005 (We decided a the best count was 0.005 m/div, in portrait view. 
ΔEy for m_{BEST }= 18 (N) (see discussion just above table.) 
ΔEy for m_{MAX }= 18 (N) (see discussion just above table.) 
ΔEy for m_{M}_{IN }= 18 (N) (see discussion just above table.) 

change in Δy for m_{BEST }= 0.475 
change in Δy for m_{MAX }= 0.440

change in Δy for m_{MIN }= 0.525 

m_{MAX}  (0.195  0.245)//[ 8  (10)] = 0.024444.  
m_{MIN}  (0.240  0.285)//[ 8  (10)] = 0.02916666.  
(m_{MAX} m_{MIN })/2  [0.024444  ( 0.02916666 )]/2 = 0.0023613  
LCy/(change in Δy)  0.005/18 = 0.000277777777  
m_{BEST} 


Δm 
Choose which ever is larger, (m_{MAX} m_{MIN })/2 or LCy/(change in Δy); in case, Δm =0.0023618 

e/M  2*m_{BEST}/(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 } = (0.0023618)/(0.02638888)*(1.74472x10 ^{11} C/kg ) = 0.1561x10 ^{11} C/kg 
E_{y} (N/C)  change in y 
12 
SEE TABLE ABOVE FOR VALUES; WHAT REALLY IS IMPORTANT IS THE COMPUTATION IN THE BOX BELOW. 
8  
4  
4  
8  
12 
Comparison 


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 = 1.75882012 (15) x 10 ^{11} C/kg . *Look up the accepted values of e and M on internet or in the textbook. 
SHOW WORK
Question 5: WITH EXCEL, MEASURE e/M = ELECTRON CHARGE TO MASS RATIO:
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.
E_{y} (N/C)  change in y 
12  
8  
4  
4  
8  
12 