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Off-axis electric field of a ring of chargeFredy R. Zypmana�
Department of Physics, Yeshiva University, New York, New York 10033-3201
�Received 29 July 2005; accepted 11 November 2005�
We consider the electric field produced by a charged ring and develop analytical expressions for theelectric field based on intuition developed from numerical experiments. Our solution involves theapproximation of elliptic integrals. Problems are suggested for an arbitrarily charged ring. © 2006American Association of Physics Teachers.
�DOI: 10.1119/1.2149869�
I. INTRODUCTION
The use of numerical methods in physics courses is al-ready a mature practice. This use is true in particular in elec-tricity and magnetism, where standard textbooks have incor-porated new chapters,1 sections,2–4 examples,5 andproblems.6 Many articles have introduced instructional ma-terial using numerical methods.7–16
We consider the problem of finding the electrostatic poten-tial in all space produced by a charged ring. In most analyti-cal approaches to this problem, the solution is found onlyalong the symmetry axis of the ring, where the integrals canbe expressed in terms of elementary functions. The off-axisproblem is avoided because it involves special functions thatare unfamiliar to most undergraduates. We surmount this dif-ficulty by plotting the relevant numerical integrals andthrough trial and error develop intuition about the expectedform of the solution. The expected form is used to constructa hypothesis about the form of the solution. We test thishypothesis and propose simple, useful formulas for the elec-trostatic field.
II. FIELD PRODUCED BY A UNIFORMLYCHARGED RING
Figure 1 shows the system of interest. A charged ring ofradius a rests in the xy plane with its center at point 0. Ageneric source point r� is parametrized by the angle � that it
makes with i, a unit vector:
r� = a cos �i + a sin � j . �1�
The observation point r= �x ,y ,z� is located anywhere inspace. Due to axial symmetry, we do not lose generality bycalculating the field at r= �x ,0 ,z�. We write the distance to
the origin as r=a� and let � be the angle between r and k:
r = a� sin �i + a� cos �k . �2�
We let Q be the total charge in the ring and write the electric
field as295 Am. J. Phys. 74 �4�, April 2006 http://aapt.org/ajp
E =1
4��0
Q
2�a�
Ring
r − r�
�r − r��3dr� �3a�
=1
4��0
Q
2�a
1
a�
0
2� �� sin � − cos ��i − sin � j + � cos �k
�1 + �2 − 2� sin � cos ��3/2 d� .
�3b�
The integral in the j direction vanishes as expected because atest charge at r does not sense a torque along the symmetryaxis. The other integrals are functions of � only throughcos �. Because cos � is an even function with respect to �=�, we can make the substitution �0
2�→2�0�. The compo-
nents of the electric field are
Ex =1
4��0
Q
�a2� � sin �
�1 + �2�3/2 f1��� −1
�1 + �2�3/2 f2��� ,
�4a�
Ez =1
4��0
Q
�a2� � cos �
�1 + �2�3/2 f1��� , �4b�
where we have introduced
� =2�
1 + �2 sin � , �5�
f1��� = �0
� d�
�1 − � cos ��3/2 , �6�
f2��� = �0
� cos � d�
�1 − � cos ��3/2 . �7�
At this stage, the expressions have been reduced as muchas possible. Further advance requires evaluating the ellipticalintegrals in Eqs. �6� and �7�. Students’ unfamiliarity withthese functions justifies a numerical approach. The goal ofthis approach is not to achieve high accuracy, but to obtainan approximation that will bring insight into the behavior ofthe electric field.
The first step is to plot f1��� and f2���. To this end, weuse MATHEMATICA because our school has a site license. Fig-ure 2 shows the first three statements of the program. Thefunction f1��� is plotted by creating a list of pairs � , f1����.Note that the physical restrictions 0���� and ��0 bound� to the interval �0, 1�. From the plot in Fig. 3 we see thatthe function looks smooth for �1 and appears to diverge at
�=1. This behavior is reasonable because we expect the295© 2006 American Association of Physics Teachers
magnitude of the field to be finite at all points in space withthe exception of points on the ring that are at �=1 and �=� /2, corresponding to �=1.
We next study numerically the nature of the divergence.Given that the field of a line charge diverges as the inversedistance and as the inverse distance squared for a pointcharge, we guess the divergence to be a power law. Thus, wenumerically construct the quantity �1−��f1��� for various. Figure 3 shows plots for =0.9, 1.1, and 1.0. We see thatfor =0.9, the divergence is still present. For =1.1,�1−��f1��� goes to zero, indicating that is smaller than1.1. On the other hand, for =1.0, �1−��f1��� remainsfinite for all values of �. Therefore, this evidence suggeststhat f1��� diverges as 1/ �1−��.
We now turn to the integral itself. Although we cannotwrite the integral f1��� in terms of elementary functions ingeneral, we can do so for ��1. We notice that when ��1,most of the contribution to the integral comes from theneighborhood of �=0. Figure 4 shows the code in which wedefine the integrand g����. We plot g���� vs � in the interval�0,� /4� for 0.95���0.99. We see that as �→1, the inte-gral is dominated by values of � in �0,0.3 rad�. This domi-
Fig. 1. The charged ring produces an electric field at the point r.
Fig. 2. The definition, evaluation, and plot of f1���.
296 Am. J. Phys., Vol. 74, No. 4, April 2006
nance suggests using 1−�2 /2 instead of cos �, which is ac-curate within 0.12% in the interval �0,0.4 rad�. Thus weconsider,
lim�→1
f1��� = �0
� d�
1 − ��1 −�2
2��3/2 , �8�
which equals
�2�
�1 − ���2 + ��2 − 2��.
Because this expression holds only for �→1, we have
lim�→1
f1��� =�2
1 − �, �9�
thus confirming our hypothesis. Note that the coefficient �2is in agreement with the graphical behavior of �1−��f1����1.4 in Fig. 3 for ��1.
The 1/ �1−�� dependence in Eq. �9� is due to the fact thatclose to the ring, the electric field of the ring is indistinguish-able from that of a line of charge. For a line of charge, it iswell known that the field behaves as a / �a−r�=1/ �1−���1/ �1−�� when r�a and ��� /2, thus justifying the di-vergent behavior here.
We now obtain the behavior of f2���. For ��1, f2���should also diverge as �2/ �1−�� because the factor −�2 /2for the expansion of cos � in the numerator does not contrib-ute appreciably to the integral for ��0. We check this state-ment numerically by plotting ��1−�� /�2�f2��� in Fig. 5,where we see that this combination remains finite for allvalues of �.
In our search for simple expressions we refer to Figs. 3and 5 where we notice that the functions ��1−�� /�2�f1���and ��1−�� /�2�f2��� behave linearly with �. On this basis,we fit them to linear functions. If high accuracy were re-quired, we could use more general polynomials, but that isnot our purpose here. We perform the approximations byconsidering the values of the functions at the end points,namely,
f1�0� = �,1 − �
�2f1��� → 1 for � → 1,
�10�
f2�0� = 0,1 − �
�2f2��� → 1 for � → 1.
Thus,
�1 − ��f1��� � � − �� − �2�� , �11a�
�1 − ��f2��� � �2� . �11b�
If we use Eq. �11� in Eqs. �4�–�7�, we arrive at explicit ex-
pressions for the electric field,296Fredy R. Zypman
297 297Am. J. Phys., Vol. 74, No. 4, April 2006 Fredy R. Zypman
Fig. 4. Integrand of the function of interest for � close to 1.
�
Fig. 5. Plot of ��1−�� / 2�f2���. The plot shows that the function remains finite for all values of �.298 298Am. J. Phys., Vol. 74, No. 4, April 2006 Fredy R. Zypman
Ex =1
4��0
Q
�a2� � sin �
�1 + �2�3/2
� −2�� − �2��
1 + �2 sin �
1 −2�
1 + �2 sin �
−1
�1 + �2�3/2
2�2�
1 + �2 sin �
1 −2�
1 + �2 sin �� , �12a�
Ez =1
4��0
Q
�a2� � cos �
�1 + �2�3/2
� −2�� − �2��
1 + �2 sin �
1 −2�
1 + �2 sin � � .
�12b�
We check Eq. �12� in the limiting cases �=0 and �→�. Inthe first case we should recover the solution for the electricfield along the symmetry axis. In the second case we shouldobtain the Coulomb field as the leading term. We set �=0and �=0 in Eq. �12� and correctly obtain,17
Ex = 0, �13a�
Ez =1
4��0
Q
�a2� �
�1 + �2�3/2� =Q
4��0
r
�a2 + r2�3/2 .
�13b�
We next let �→� in Eq. �12� and obtain
Ex →1
4��0
Q
r2 sin��� , �14a�
Ez →1
4��0
Q
r2 cos��� , �14b�
which are the projections along the axes of the Coulombfield.
Fig. 6. Electric field in the xz plane.
In Fig. 6 we show a plot of the electric field given by Eq.
299 Am. J. Phys., Vol. 74, No. 4, April 2006
�12�. The magnitude of an arrow has been set to a constantvalue to emphasize the local direction of the field. On alinear scale most arrows would be too small to be displayeddue to the large values of the field near the ring at the point�a ,0�. Figure 6 clearly shows the divergence at the ring, andthe axial direction close to the axis of symmetry.
What is the importance of having off-axis information?The electric field lines give the trajectories of charged par-ticles in that region. This information has been used exten-sively in the design of devices to confine charged particles�plasmas� in small volumes18 and in the discrimination ofparticles according to mass �mass spectrometers�.19 Electro-static problems with axial symmetry can be handled by sum-mation of charged rings. Relevant problems are related to thecalculations of tip-sample forces in atomic forcemicroscopy.20,21 Particle guns, electrostatic velocity selec-tors, and electrostatic ion thruster are described in Ref. 22.Electrostatic lenses are discussed in Ref. 23.
III. GENERALIZATIONS
We next consider a ring with a nonuniform charge density.This problem can be effectively used as a homework project.
The first difficulty is that, in general, the electric field doesnot have cylindrical symmetry. In this case, we need to workwith expressions as in Eq. �3� with a linear charge densityfunction, ����. Although not all planes through z are equiva-lent, we will analyze the problem in the xz plane. There is noloss of generality because the analysis for any plane throughz is the same if the origin of � is properly chosen.
Consider ���� defined in −���� and expand it around�=0. As discussed in Sec. II, when ��1 we can assume�������0� for ��0. Therefore, we expect the divergencesto behave like 1/ �1−�� in this case. We then proceed asbefore, multiply the original integrals by 1−�, and then fitthese finite functions to simple polynomials.
Information about the field from a nonuniformly chargedring has a variety of applications. The quadrupolar chargedistribution on a ring has been used in high precision mea-surements of properties of elementary particles.24 Gravita-tional fields, which also satisfy Poisson’s equation,25 are use-ful for gravitational propulsion. A knowledge of thegravitational field produced by the rings of Saturn requiresthe solution of the electrostatic ring with nonuniform charge.Another example is precision measurements of theAharonov-Bohm effect. And to study the interplay betweenthe magnetic, electrostatic, and Aharonov-Bohm effects, anano-ring with nonuniform potential was constructed.26
IV. DISCUSSION
We solved the electrostatic problem of a charged ring byfinding simple solutions that retain the character of the field’sbehavior. Similar integrals appear for the magnetic field pro-duced by a current ring. This problem is of particular interestbecause it forms the basis for understanding the parametersof Helmholtz coils. For the same reason as in the electro-static case, the off-axis solution is usually avoided. With theexpression provided in this paper, further questions can beasked regarding the homogeneity and strength of the mag-netic field from the symmetry axis.
The approach used to investigate the nature of the diver-gence in this paper could be used in a large variety of prob-
lems, even those that have nonpower law divergences. The299Fredy R. Zypman
idea is to build student intuition about the nature of the di-vergence �if it exists� via numerical exploration, which mayinclude plotting the functions with nonlinear axes.
ACKNOWLEDGMENT
This work was supported by the Research Corporationthrough Grant No. CC5786.
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