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Unformatted text preview: ltage (103 V)
1.39
2.21
2.78
3.11
4.22
? Plate separation (mm)
11.1
20.0
24.0
28.1
35.1
50.0 Written Response — 15%
2. • Provide a graph of the balancing voltage as a function of the plate separation, with
the manipulated variable on the horizontal axis.
• Calculate the slope of the graph, and describe the physical quantity or quantities
that this slope represents.
• Using the slope, or another suitable averaging technique, determine the magnitude
of the charge on the suspended mass.
• Determine the balancing voltage required when the plates are separated by 50.0 mm.
Clearly communicate your understanding of the physics principles that you
are using to solve this question. You may communicate this understanding
mathematically, graphically, and/or with written statements. 11 Sample Solution
• Provide a graph of the balancing voltage as a function of the plate separation, with
the manipulated variable on the horizontal axis.
Voltage as a Function of Plate Separation
6.0 5.0 4.0
Voltage (103 V) 3.0 2.0 1.0 0.0
0 10 20
30
40
Plate separation (10–3 m) 50 • Calculate the slope of the graph and describe the physical quantity or quantities
this slope represents.
slope =
=
= rise/run
(4.4 × 103 − 1.4 × 103 ) V
(38 × 103 − 12 × 10−3 ) m
1.15 × 105 V/m or consistent with graph The slope is the electric field between the plates
or
The slope is the gravitational force (weight) of the mass divided by the charge on
the object 12 • Using the slope, or another suitable averaging technique, determine the magnitude
of the charge on the suspended mass.
Balanced forces gives:
Fg = mg and Fe = q E and E = V
d Fg = Fe
qV
d
mgd
q=
V mg = Method 1 Using Slope
V
slope =
d
mg
therefore, q =
slope
(3.8 × 10 −15 kg)(9.81 m/s 2 )
q=
(1.15 × 10 5 V/m) q = 3.2 × 10−19 C
Method 2 Data Averaging q1 = (3.8 ×10 −15 kg)(9.81 m/s 2 ) × 11.1×10 −3 m
= 2.98 ×10−19 C
3
1.39 ×10 V q2 = (3.8 × 10 −15 kg)(9.81 m/s 2 ) × 20.0 × 10 −3 m
= 3.37 × 10 −19 C
3
2.21 × 10 V q3 = (3.8 ×10 −15 kg)(9.81 m/s 2 ) × 24.0 ×10 −3 m
= 3.22 ×10 −19 C
3
2.78 ×10 V q4 = (3.8 × 10 −15 kg)(9.81 m/s 2 ) × 2.81 × 10 −3 m
= 3.37 × 10−19 C
3
3.11 × 10 V q5 = (3.8 × 10 −15 kg)(9.81 m/s 2 ) × 3.51 × 10−3 m
= 3.10 × 10−19 C
3
4.22 × 10 V q= Σq 1.604 ×10−18 C
=
5
n q = 3.2 ×10−19 C 13 • Determine the balancing voltage required when the plates are separated by
50.0 mm.
Method 1 Extrapolation from Graph
The balancing voltage required for a plate separation of 50.0 mm can be found by
reading up from 50.0 mm on the xaxis to the line of the graph and then left to the yaxis. The balancing voltage is 5.8 × 103 V
Method 2 Numerical Analysis
V
so V = (slope)d
d
V = (1.15 × 105 V/m)(5.0 × 10−2 m) slope = V = 5.8 × 103 V 14 Special Case – Use of Graphing Calculator
• Provide a graph of the balancing voltage as a function of the plate separation, with
the manipulated variable on the horizontal axis.
Balancing Voltage as a Function of Plate Separation
y x: [8.7, 37.5, 1]
x y: [0.9089, 4.7011, 1] xaxis is plate separation in mm (or 10 −3 m)
yaxis is balancing voltage in 10 3 V
• Calculate the slope of the graph and describe the physical quantity or quantities
this slope represents.
L1
L2 =
= plate separation, d, in mm (or 10 −3 m)
balancing voltage, V, in 103 V Using the linear regression function on the calculator
(ax + b) L1, L2, Y
gives slope = 1.16 × 105 V /m
The slope is the electric field between the plates
or The slope is the gravitational force (weight) of the mass divided by the charge on
the object
Notes
1. the definition (and units) of L1 and L2 and the order they are used in the linear
regression must be consistent • Using the slope, or another suitable averaging technique, determine the magnitude
of the charge on the suspended mass.
Same as given in regular sample solution
• Determine the balancing voltage required when the plates are separated by
50.0 mm.
By extending window to include x = 50.0 mm and then using the trace function,
the balancing voltage is 5.8 × 103 V
15...
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 Fall '09
 Quinlan
 Physics

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