FALL 2009 – 2010
PHYS-102 I5
Equipotential Lines
NEBRASKA WESLEYAN UNIVERSITY
NAME: ___________________________
SECTION:
Numerical Solution for the Potential Surface for 2D
Charged Conductors. Since the electrostatic potential V is
a scalar quantity, a two-dimensional (2D) potential field can be
plotted as a surface. In order to generate a 2D potential field, we use
charged conductive bodies that are long and of constant crosssection, such as rods and bars that are oriented perpendicularly to the
plane of interest, as shown in the accompanying figure. Recall that
for such bodies, the electric field lines lie entirely in the plane of
interest because it is a plane of symmetry.
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8am
10am
1pm
3pm
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PLANE OF
INTEREST
+
−
We wish to compute and visualize the potential in the plane of
interest. Unfortunately, but no closed form mathematical formula can
be written to express the potential at each point in space. This is
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because a conductive body (whose excess charge lies on its surface)
possesses a charge distribution that is complicated and not
predictable ahead of time. Fortunately, there is a simple numerical
formula that can be easily derived using Gauss’s Law and
implemented using an Excel spreadsheet. Because the solution is
numerical, it involves a discrete number of finite area elements within the plane of interest, which we will choose to be
squares of edge length x. The key to applying Gauss’s Law is in the selection of the Gaussian surface shown in the
accompanying figure. The surface is a rectangular solid whose long axis is perpendicular to the plane of interest, and whose
PLANE OF
INTEREST
BACK (B)
GAUSSIAN
SURFACE
y
TOP
LEFT (L)
RIGHT
FRONT
2
h
3
P
1
x
RIGHT (R)
4
FRONT (F)
PLANE OF
INTEREST
x
other axes are along the x and y coordinate axes. Viewing the plane of interest along a line perpendicular to it, the Gaussian
surface is drawn as a solid square, of size equal to that of an area element. Notice that there are other adjacent area elements
of equal size shown as dashed squares. Each area element has a center, marked with a dot. Each dot can be thought of as the
mathematical point at which the potential is to be determined. The Gaussian surface is assumed to lie only at elements
representing regions with no net charge enclosed. Elements associated with conductive bodies have zero electric field
throughout, and therefore have V specified as a constant value beforehand.
Explain why conductors have a constant electrostatic potential V inside them.
Explain why the total electric flux FE passing out of the Gaussian surface discussed above would be zero.
DEAN SIEGLAFF
CREATED: 30 JAN 2010
MODIFIED:
1 of 4
PHYS-102 I5
Equipotential Lines
FALL 2009 – 2010
NEBRASKA WESLEYAN UNIVERSITY
With the help of your instructor, express the electric flux E passing through each of the six faces of the Gaussian
surface. Label the electric field at the left face as EL, for example.
FACE
EXPRESSION FOR ELECTRIC FLUX
EXPLANATION / NOTES
RIGHT
LEFT
BACK
FRONT
TOP
BOTTOM
Express the required components of the electric field for each of the four faces in terms of the potentials at the dots and the
distances between them x. Note that we assume x is small, so that the electric field is assumed constant over that distance.
EXPRESSION IN TERMS
FACE
ELECTRIC FIELD
COMPONENT SYMBOL
RIGHT
ERx
LEFT
ELx
BACK
EBy
FRONT
EFy
OF POTENTIAL V AND
DISTANCE x
EXPLANATION / NOTES
Put all of the above information together into Gauss’s Law, and find the simple condition upon the potential at point P, VP, in
terms of the potential at each of its four nearest neighbors.
VP =
DEAN SIEGLAFF
2 of 4
PHYS-102 I5
Equipotential Lines
NEBRASKA WESLEYAN UNIVERSITY
FALL 2009 – 2010
Applying the Numerical Solution for the Potential Surface. Excel spreadsheet is the perfect tool for implementing
the numerical solution found above. With Excel, we will make the calculations, and then visualize the result as a
surface plot and a contour plot.
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Launch Excel workbook “PHYS-102 I5 V Surface Starting Point.xls.”
“Enable iterative calculation” found in Office Button (upper left) → Excel Options → Formulas, with “Maximum
iterations” of 200 and “Maximum change” of 0.00001.
Enter the formula “=AVERAGE(R[-1]C,RC[-1],R[1]C,RC[1])” into R2C2.
Fill that formula to R2C49. Note: Do not effect the boundary cells, which have the constant value “0” in them.
Select R2C2 to R2C49, and fill those formulas down to R49C2 to R49C49. Note: Do not effect the boundary
cells.
Enter “10” into cells R14C24 through R16C26. Note: Format those cells with a fill color for ease of location.
Allow the application to iterate through the maximum number of iterations as entries are made.
Enter “−10” into cells R34C24 through R36C26. Note: Format those cells with a fill color for ease of location.
Strike the F9 key (or increase the maximum iterations) until the array of values relaxes to the correct solution (Excel
will no longer iterate the sheet).
View the surface plot by clicking the “Surface” tab, and edit the “3D View” (obtained by right clicking on the plot)
to edit the viewing angle and perspective.
We have created the potential surface for a physical dipole charge configuration. On the surface plot, lines of contour are
marked with different colored boundaries. These contour lines are also called lines of equipotential, because the potential is
constant along those boundaries.
Explain why the electric field always points “downhill” on a potential surface. Hint: Use the “path integration” argument.
Along which direction is the maximum change in potential?
Switch to the EM Field application, which makes drawing equipotentials and field lines much simpler.
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Create a physical dipole using two equal but opposite rod charges (1 and –1).
Draw 8 electric field lines emanating from each charge.
Draw a series of equipotential lines on the system.
What is the relationship between the field lines and the equipotential lines every point they cross one another? Why?
Where is the potential zero (excepting the outer boundary) on the surface between the two charged rods? If the potential is
zero, does that mean the electric field is zero? Explain.
DEAN SIEGLAFF
3 of 4
PHYS-102 I5
Equipotential Lines
NEBRASKA WESLEYAN UNIVERSITY
FALL 2009 – 2010
Two Equal Charges.
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Save your Excel workbook containing the dipole potential surface solution, because you will want to look at it later.
Save it to the desktop for convenience.
Change the “–10” values to “10,” and relax the array using the F9 key.
Where is the electric field zero on the surface between the two charged rods? Explain in terms of Coulomb’s Law and in
terms of the slope of the surface. Hint: Look for the “saddle” point. It is a position of zero slope for the surface.
Is the potential zero at that location?
If the electric field is zero, does that mean the potential is zero? Explain.
Quadrupole Potential Surface.
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Save your Excel workbook containing the two equal charge potential surface solution.
Delete the existing charged rods from the data sheet by filling formula cells over them.
Enter “10” into cells R14C14 through R16C16 and cells R34C34 through R36C36. Note: Format those cells with
a fill color for ease of location. Allow the application to iterate through the maximum number of iterations as
entries are made.
Enter “−10” into cells R34C14 through R36C16 and cells R14C34 through R16C36. Note: Format those cells
with a fill color for ease of location.
Strike the F9 key (or increase the maximum iterations) until the array of values relaxes to the correct solution (Excel
will no longer iterate the sheet).
Build the same charge configuration in EM Field using +1 and -1 rod charges. Map the electric field and the lines of
equipotential.
Where is the potential zero? Where is the electric field zero? Can the electric field and the potential both be zero at the same
location? Explain.
At points where the electric field is zero, how do the electric field lines behave? How do the equipotential lines behave? Are
field lines ever supposed to cross? Explain.
DEAN SIEGLAFF
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