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Pre-Lab Quiz / PHYS 224
Electric Field and Electric Potential (B)
Your Name _________________________Lab Section _____________
1. What do you investigate in this lab?
2. For two concentric conducting rings, the inner radius of the larger
ring is 12.0 cm and the outer radius of the smaller ring is 4.0 cm. The
electric potential of the inner ring is 0 V and that of the outer ring
is 10 V. Calculate the electric potential half way between the two
rings, i.e., 8.0 cm away from the center of the rings.
3. In the following figure, the electric potential of the inner
conducting ring is 0 and the electric potential of the outer conducting
ring is β5.0 V. The two rings are concentric. Draw four equipotential
lines with 1 V-difference between two neighboring lines and label each
line with the corresponding voltage. Also draw four electric field lines
in each figure.
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Instructorβs Lab Manual / PHYS 224
Electric Field and Electric Potential (B)
Name _____________________________ Lab Section _____________
Objectives In this lab, you study the electric field between two coaxial metal
rings, map the equipotentials, determine the electric field, and plot
the electric field lines. The configuration resembles closely to a
uniaxial electric field.
Background
In electrostatics, the electric field (π¬π¬οΏ½οΏ½β ) and the electric
potential (ππ) are two interchangeable physical quantities for describing
electric phenomena, although the former is a vector and the latter is a
scalar.
When placing a test electric charge q at a point P in space, it
experiences an electric force that is given by
πποΏ½οΏ½β = ππ π¬π¬οΏ½οΏ½β , (1)
where π¬π¬οΏ½οΏ½β is the electric field at point ππ. When the test charge moves
from point ππ to point ππ, the electric force on the charge is always
proportional to the electric field along the pathway. The total work
done by the electric force can be obtained by integrating the work along
the pathway. Because the electrostatic force is conservative, the total
work depends only on the starting and ending points of the pathway,
regardless on the specific pathway the charge traverses between those
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points. One can thus introduce the electric potential difference between
the two points, which is related to the total work (βππ) done by the
electric force, as described by equation
βππ = βππ βππ = βππ οΏ½ππππ β πππποΏ½ (2)
Therefore, if one maps out the electric field in a certain region
of space, one can calculate the electric potential in that region (apart
from a common constant), and vice versa.
As a vector quantity, the electric field at a point must be
described by its magnitude and direction, altogether requiring three
numbers. To visualize π¬π¬οΏ½οΏ½β in space, one normally draws electric field
lines (each line with an arrow to designate its direction) in space.
The direction of π¬π¬οΏ½οΏ½β at a point is parallel to the tangent of the electric
field line passing through that point.
As a scalar quantity, the electric potential at any point requires
only one number to describe. As such, to plot the electric potential
in space, one can simply find and draw equipotential surfaces (often
referred simply as equipotentials). For every point of an equipotential
surface, the electric potential is the same.
Clearly, it is much easier to map and plot electric potential in
space. Normally, one draws a series of equipotentials with the same
potential difference between any two neighboring equipotentials. To
obtain the electric field from the plot of equipotentials, one uses the
following rules. At any point on an equipotential surface, the electric
field is always perpendicular to the equipotential surface, and the
electric field points from a high potential to a low potential. The
magnitude of the electric field is inversely proportional to the distance
between two neighboring equipotentials. If the potential difference
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between two infinitely close points π΄π΄ and π΅π΅ is βππ = πππ΅π΅ β πππ΄π΄ and their
distance is βππ, the component of the electric field projected in the
direction pointing from A to B is given by οΏ½π¬π¬οΏ½οΏ½β οΏ½ = βππβπποΏ½ .
A uniaxial electric field can be induced between two concentric
conducting cylindrical shells ππ and ππ with their longitudinal lengths
much larger compared to their separation. Assume the inner radius of
the larger ring as ππππ and the outer radius of the smaller ring as ππππ.
When an electric potential difference βππππ = ππππ β ππππ is established
between the two rings, at any point between the rings the electric field
points radially and its magnitude depends only on the distance between
the point and the symmetry axis of the rings. By using calculus between
the rings and at a distance of ππ from the symmetry axis of the rings,
the magnitude of the electric field is given by
πΈπΈ = πΆπΆππ (3)
where the constant πΆπΆ = (ππππ β ππππ)ππππ οΏ½ππππ
πππποΏ½οΏ½ . Thus, equipotentials are also
concentric rings. The potential difference between two equipotentials is
however not linearly proportional to their distance. Between the rings
and at a distance of r from the symmetry axis of the rings, the electric
potential is given by
ππ(ππ) = ππππ + πΆπΆ ππππ οΏ½ πππππποΏ½. (4)
The S.I. units for the above equations are: E in volts-per-meter
(V/m), d or r in meter (m), and V in volt (V).
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EXPERIMENT
Apparatus
Fill up with water the rectangular tray to a height of 0.5-inch.
Place the circular graph paper underneath the tray. Two metal rings are
used as electrodes to study uniaxially symmetric electric field. The
outer diameter of the smaller ring is 1.9 in and the inner diameter of
the larger ring is 6.0 in. Place the two metal rings inside the tray
and align their centers with the center on the graph paper.
Because the longitudinal lengths of the two metal rings are short
compared to their separation, the electric field between the two metal
bars cannot be considered exactly as uniaxial. Therefore, your
measurements in this lab may not completely follow Equations (3) and
(4).
Procedures Figure 1 displays the circular graph paper (not to scale) to be
used in this experiment. We will use polar coordinates (ππ, ΞΈ) to identify
specific points. Note: the unit for the ππ coordinate is inch.
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1. Set up the circuit (Figures 2-4)
Keep the power switch of the DC power supply in the off position.
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Follow Figure 2 to start the setup of the electrical circuit. i)
Place a point probe with a round tip (P1) over the inner metal ring.
Connect this point probe (P1) to the negative (ground) port of the DC
power supply. ii) Place a point probe with a round tip (P2) over the
outer metal ring. Connect this point probe (P2) to the positive terminal
of the DC power supply.
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As shown in Figure 3, iii) connect the point probe P3 to the common
port of Voltmeter A. Connect the handheld probe P4 to the voltage port
of Voltmeter A. Note: the sensitivity level of the voltmeter should be
set to 20 V (when you are asked to use it).
As shown in Figure 4, iv) connect probe P3 to the voltage input
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port of Voltmeter B. Connect the common port of Voltmeter B to probe
P1. Note: the sensitivity level of the voltmeter should be set to 20 V
(when you are asked to use it)
The point probes P1 and P2 will be used to set the potential across
the two metal rings, which will be provided by the DC power supply. The
point probe (P3) and the hand-held point probe (P4) will be used to map
potential through voltmeter A in the region between the rings. The tips
P1 and P2 should be always under water and in contact with the respective
metal ring. Make sure the two rings do not move during the experiment.
Also P3 and P4 probes should be under water when collecting data.
Ask your TA to check the circuit!
2. Investigate Electric Potential-versus-Distance
Turn on the DC power supply. Set the output voltage of the DC power
supply to 12 V.
Make sure Voltmeter B is off. Set the voltage level of Voltmeter A
to 20 V and turn it on. Place the tip of probe P3 at point (ππ = 1.5,
ΞΈ = 0Β°). Place the tip of probe P4 at points with ΞΈ = 0Β° and in turn with
ππβ² = 1.7, 1.9, 2.1, 2.3, and 2.5. Read the corresponding potential difference
(β ππ) between the P4 and P3 probes from Voltmeter A. Record them in Table
1. Calculate β ππ /(ππ β ππβ²) and record the ratio values in Table 1. (Note:
1 inch = 0.0254 m.)
TABLE 1
ππβ²
(inch)
Potential difference βππ
(V)
β ππ /(ππ β ππβ²)
(V/m)
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1.7
1.9
2.1
2.3
2.5
3. Map the first equipotential
Turn off Voltmeter A. Set voltage level of Voltmeter B to 20 V and
turn it on. Make sure the tip of probe P3 is at the point (ππ = 1.5, ΞΈ = 0Β°).
Read the potential at this point from Voltmeter B. Mark the point in
Figure 1. Record it in Table 2.
Turn off Voltmeter B and turn on Voltmeter A. Move the tip probe
P4 along the ΞΈ = 90Β° line until Voltmeter A reads 0 V. This means that
the tips of P3 and P4 are placed on the same equipotential. Read out the
ππ coordinate of the tip of P4 and mark the point in Figure 1. Record it
in Table 2.
Repeat the measurement by moving the tip of P4 respectively along
the ΞΈ = 180Β° and 270Β° lines until Voltmeter A reads 0 V. Read out the
corresponding ππ coordinates of the tip of P4 and mark the corresponding
points in Figure 1. Record them in Table 2. Draw a curve through the
four points.
TABLE 2
Tip of P3
Measured
potential
(V)
ππ for
ΞΈ = 90Β°
ππ for
ΞΈ = 180Β°
ππ for
ΞΈ = 270Β°
Average
ππ
Calculated
potential
(V)
(ππ = 1.5, ΞΈ = 0Β°)
(ππ = 1.7, ΞΈ = 0Β°)
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(ππ = 1.9, ΞΈ = 0Β°)
(ππ = 2.1, ΞΈ = 0Β°)
(ππ = 2.3, ΞΈ = 0Β°)
(ππ = 2.5, ΞΈ = 0Β°)
4. Map five more equipotentials
Repeat step 3 to map four more equipotentials by moving the tip of
probe P3 respectively to points (ππ = 1.7 , ΞΈ = 0Β°), (ππ = 1.9 , ΞΈ = 0Β°), (ππ =
2.1 , ΞΈ = 0Β°), (ππ = 2.3 , ΞΈ = 0Β°), and (ππ = 2.5 , ΞΈ = 0Β°).
Questions
1. For the measurement by step 2, should the measured β ππ be linearly
proportional to β ππ?
2. Use an arrow to mark the directions of the electric field at (ππ =
2.0, ΞΈ = 45Β°) and (ππ = 2, ΞΈ = 225Β°) in Figure 1.
3. When using Equation (4) to describe the measured potential-versus-
radius, what are the values of ππππ, ππππ, ππππ, ππππ , ππππππ πΆπΆ?
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Analysis Calculate the averaged ππ coordinate for the four points on each
equipotential. Use the averaged ππ coordinates and Equation (4) to
calculate the corresponding potential. Record them in Table 2 and compare
with the measured corresponding potentials.
Use the averaged ππ coordinates and the measured corresponding
potentials to calculate the electric field values οΏ½π¬π¬οΏ½οΏ½β οΏ½ = βππβπποΏ½ between
every two neighboring equipotentials. (Note: 1 inch= 0.0254 m.)
Between the 1st and 2nd: οΏ½π¬π¬οΏ½οΏ½β οΏ½ =
Between the 2nd and 3rd: οΏ½π¬π¬οΏ½οΏ½β οΏ½ =
Between the 3rd and 4th: οΏ½π¬π¬οΏ½οΏ½β οΏ½ =
Between the 4th and 5th: οΏ½π¬π¬οΏ½οΏ½β οΏ½=
Between the 5th and 6th: οΏ½π¬π¬οΏ½οΏ½β οΏ½
In Figure 1, draw four electric field lines.