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The potential, electric field and surface charges for a resistive long straightstrip carrying a steady current

J. A. Hernandesa)

Instituto de Fsica Gleb Wataghin, Universidade Estadual de Campinas-Unicamp, 13083-970 Campinas,Sao Paulo, Brazil

A. K. T. Assisb)

Institut fur Geschichte der Naturwissenschaften, Universitat Hamburg, Bundesstrasse 55,D-20146 Hamburg, Germany

Received 20 March 2002; accepted 24 March 2003

We consider a long resistive straight strip carrying a constant current and calculate the potential and

electric field everywhere in space and the density of surface charges along the strip. We compare

these calculations with experimental results. 2003 American Association of Physics Teachers.

DOI: 10.1119/1.1574319

I. THE PROBLEM

Recently there has been renewed interest in the electricfield outside stationary resistive conductors carrying a con-stant current.1 7 We consider a case that has not been treatedin the literature, namely, a constant current flowing uni-

formly over the surface of a stationary and resistive straightstrip. Our goal is to calculate the potential and electricfield E everywhere in space and the surface charge distribu-tion along the strip that creates this electric field.

We consider a strip in the y0 plane localized in the

region axa and z , such that a0. The

medium around the strip is taken to be air or vacuum. The

constant currentIflows uniformly along the positive z direc-

tion with a surface current density given by KI z/2a seeFig. 1. By Ohms law this uniform current distribution isrelated to a spatially constant electric field on the surface ofthe strip. In the steady state this electric field can be related

to the potential by E. This relation means that along

the strip the potential is a linear function of z and indepen-dent of x. The problem can then be solved by finding the

solution of Laplaces equation 20 in empty space andapplying the boundary conditions.

II. THE SOLUTION

Due to the symmetry of the problem, it is convenient to

use elliptic-cylindrical coordinates ,,z ).8 These variables

can take the following values: 0 , 02, and

z . The relation between Cartesian (x, y , z) andelliptic-cylindrical coordinates is given by

xacosh cos , 1a

yasinh sin , 1b

zz , 1c

where a is the constant semi-thickness of the strip. The in-

verse relations are given by

tanh1x2y 2a 2

2x2 , 2a

tan1a2x2y 2

2x 2 , 2b

zz , 2c

where (x2y 2a 2)24a 2x2.Laplaces equation in this coordinate system is given by

2 1a2cosh2 cos2

2

2

2

2

2

z 20.

3

A solution of Eq. 3 can be obtained by using separation ofvariables in the form (,,z)H()()Z(z):

H23a2 cosh2 H0, 4a

23a2 cos2 0, 4b

Z3Z0, 4c

where 2 and 3 are constants.For a long strip being considered here, it is possible to

neglect boundary effects near z. It has already been

proved that in this case the potential must be a linear func-

tion of z, not only over the strip, but also over all space.9

This condition means that 30. There are then two pos-

sible solutions for . If 20, then C1C2; if

20, then C3sin2C4cos2, whereC1 to C4are constants. Along the strip we have y0, and x 2a2,

which means that a 2x2, 0, and

tan1(a 2x 2)/x2. Because the potential does not dependonx along the strip, this independence means that the poten-

tial will not depend on as well. Thus a nontrivial solution

for can only exist if20, C20, and constant for

all . The solution for H with 230 will be then alinear function of. The general solution of the problem isthen given by

A 1A 2A 3zA 4

A 1tanh1x2y 2a 2

2x2 A 2A 3zA 4. 5

The electric field E takes the following form:

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EA 1xx 2y 2a 2x&

x

xy&

x2y 2a 2y A 3zA 4

A 3A 1tanh1x2y 2a 2

2x 2 A 2 z. 6

To find the surface charge density, we utilized the approxi-

mation close to the strip ( xa andy a):

EA 1 xy a 2x23/2x yy a 2x 2y A3zA 4A 3A 1tanh1 ya2x 2A 2z. 7

The surface charge density (x , z) can be obtained by the

standard procedure utilizing Gausss law SE"daQ/0 ,

where 0 is the vacuum permittivity, da is a surface areaelement pointing outward normal to the surface in each

point, and Q is the total charge inside the closed surface S.

The surface charge density is then obtained by considering

the limit in which y 0 in Eq. 7 and a small cylindrical

volume with its length much smaller then its diameter, yield-ing: 0E(y0) "yE(y0) "(y) . If we use Eq. 7,the surface charge density is found to be given by

20A1A 3zA 4

a 2x2 . 8

III. DISCUSSION

In the plane y0 the current in the strip creates a mag-

netic field B that points along the positive negative y di-

rection for x0 (x0). This magnetic field will act on the

conduction electrons moving with drift velocity vd with a

force given by qvdB see Fig. 2. This force will cause a

redistribution of charges along the x direction, with negativecharges concentrating along the center of the strip and posi-

tive charges at the extremities xa. In the steady state this

redistribution of charges will create an electric field along the

x direction,Ex, that will balance the magnetic force, namely,

qEx qvdB.We have disregarded this Hall electric field because it is

usually much smaller than the electric field giving rise to thecurrent.10 To estimate the orders of magnitude involved, it is

easier to consider the current Iflowing uniformly in a long

cylinder of length 2and radiusa along the positive z direc-

tion coinciding with the axis of this cylinder. This current

generates a cylindrical magnetic field given by at a distancera from the axis B0Ir/2a2, where 0 is the

vacuum permeability and is the unit polar vector. The

magnetic force acting on an electron of charge qe mov-

ing with drift velocity vdvdz relative to the lattice ofthe wire is given by qvdB 0evdIr/2a

2r, where ris the unit radial vector. This inward radial force will lead toan accumulation of negative charges in the interior of the

wire, which creates a radial electric field Erpointing inward.In the steady state the electric and magnetic radial forces will

balance one another, qErqvdB, yielding Er0vdIr/2a

2 r. This electric field increases linearly in-side the wire. Its maximum value close to ra is given by

Ermax

0vdI/2a. The longitudinal electric field givingrise to the current can be obtained by Ohms law, VRI,

where V is the electromotive force along the wire of resis-

tanceR . For a wire of length 2 acted on by a longitudinal

electric fieldE pointing along the z direction, this voltageis given by V2E, such that ERI/2. The ratiobetween the maximal radial electric field and the longitudinal

one is given by Ermax/E0vd/aR0vdga/2, where g

is the conductivity of the wire and is related to its resistance

by R2/ga2. We use the notation g instead of the more

standard notation for the conductivity in order to avoidconfusion with the surface charge density, which is repre-sented by .

To find the order of magnitude, we consider a copper wire(g5.7107 m and vd4103 m s1) of 1 mm diam-

eter (a5104 m). With these values in Eq. 8, we ob-

tain Ermax/E710

5, justifying our neglect of the radial

component of the electric field. Conceptually this neglect ofthe radial component can be explained by the fact that theHall electric field is small because it is due to the smallmagnetic field produced by the conducting strip, rather thana large applied magnetic field.

We now analyze some particular cases. We first consider

two limits by comparing a with the distance of the observa-

tion point rx 2y 2. If a 2r2, we have a 2y 2x 2

2x 2y 2/a 2 and y /a , such that

A 1 y a A 2 A 3zA4. 9As expected, this result coincides with Eq. 4 of Ref. 11with y 00, because only the case a

2r2 was considered

there.

On the other hand, if a 2r2, we have r2a 2

2a 2x 2/r2 and ln r/a, such that

A 1ln raA 2 A 3zA 4. 10

Fig. 1. A constant current Iflows along the z direction of a long straight

strip of length 2 and width 2a located at y0, with a surface current

density given by KIz/2a.

Fig. 2. Magnetic force FqvdB directed along the center of the strip

acting on a conduction electron moving with drifting velocity v d. This force

is due to the magnetic field B generated by the electric current I flowing

along the positive z direction.

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This result coincides with Eq. 8 of Ref. 10 with A2/A1 ln(2/a), where is the typical length of the wire or strip

being considered, with a. Note that in Ref. 10 the length

of the wire along z goes from /2 to /2, while here it

goes from to see Fig. 1.This coincidence is reason-able because Eq. 8 of Ref. 10 corresponds to the potentialoutside a long straight cylindrical wire carrying a constantcurrent. At a point far from the axis of the strip, both resultscoincide as they must.

Equation 6 indicates that there is an electric field not

only along the resistive strip carrying a constant current, butalso in the space surrounding it. Jefimenko has peformedsome experiments that show the existence of this externalelectric field. The geometry of his first experiment,12 repro-duced in plate 6 of Ref. 13, is equivalent to what has beenconsidered here: a two-dimensional conducting strip madeon a glass plate using a transparent conducting ink. To com-pare our calculations with his experimental results, we need

the values of A 2/A 1 and A 4/A 3 . We take A 2/A 13.6 and

A 4/A30. The condition A 4/A 30 corresponds to the sym-

metrical case considered by Jefimenko in which the electric

field is parallel to the conductor just outside of it at z0

zero density of surface charges at z0).

We first consider the plane orthogonal to the strip, x0. In

this case the potential reduces to

A 1tanh1 y2

y 2a 2A2 A 3zA4. 11

The lines of the electric field orthogonal to the equipotentialscan be obtained by the procedure in Ref. 14. These lines are

represented by a function such that 0. Equation11, together with the value ofobtained above, yield thevalue ofgiven by

A 1A 3z22A 1A4z

A 1A3

2 y 2

A1A 3y y2a 2 cosh1y

2a 2

a2

A1A 3

2 a 2 cosh1y 2a 2

a2

2

A2A 3

4 yy 2a 2a 2 ln y y 2a2

a . 12

A plot of Eqs. 11 and 12 is given in Fig. 3.We now consider the plane of the strip, y0. The poten-

tial reduces to

xa,0,z A 2A 3zA 4, 13

xa,0,z A 1tanh1x2a 2

x2 A 2 A3zA 4

A 1cosh1xa A 2 A 3zA 4. 14When there is no current in the strip, the potential along it

is a constant for all z. From Eq. 13 this condition implies

thatA 30. This value ofA 3 in Eqs. 5,6, and8reducesthese equations to the known electrostatic solution of a stripcharged to a constant potential.15

By a similar procedure, the lines of electric field for the

plane y0 are given by

xa,0,z A 2A 3ax, 15

xa,0,z A 1A 3z22A 1A 4z

A 1A 3

2 x2

A 1A3xx2a 2 cosh1

x

a

A 1A3

2 a2

cosh1

x

a

A 2A3

4 xx 2a2

a 2 lnxx 2a 2

a . 16

A plot of Eqs. 1316 is presented in Fig. 4. Figure 5presents the theoretical electric field lines and equipotentiallines overlaid on the experimental results of Ref. 12, wherethe lines of the electric field in the plane of the strip aremapped by spreading grass seeds above and around the two-dimensional conducting strip painted on glass plates. The

seeds are polarized in the presence of an electric field andalign themselves with it. The lines of electric field are thenobserved in analogy with iron fillings generating the lines ofmagnetic field. In Fig. 5 the electric field lines from Fig. 4are overlaid on the experimental results of Jefimenko, Fig. 1of Ref. 12 or plate 6 of Ref. 13. It should be mentioned thatthe grass seeds are dielectric bodies and themselves changethe electric fields in their vicinity, so the experimental fieldmaps cannot be exact; nevertheless, the correspondencefound here is reasonable.

The equipotential lines also were measured in Ref. 16where a rectangular hollow chamber with electrodes alumi-

Fig. 3. Equipotential lines dashed and electric field lines continuous in

thex0 plane. The bold horizontal lines represent the intersection with the

plane of the strip. We use the values A 2/A 13.6 and A 4/A 30.


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num foil for end walls and semi-conducting side wallsgraphite paper strips carrying uniform current was used.Eighty volts were applied to the electrodes and the equipo-tential lines were mapped utilizing a radioactive alpha sourceto ionize the air at the points where the field was to be mea-sured. The alpha source acquired the same potential as thefield at those points and the potential was measured with anelectronic electrometer connected to the alpha source. In Fig.6 the experimental result of Ref. 16 is superimposed on theequipotential lines calculated utilizing Eqs. 15 and 16

with A 2/A 13.0 and A 4/A 30. The agreement is not as

good as in Fig. 5 for two reasons: One reason is that ourcalculations are for a two-dimensional geometry, while theexperiment in Ref. 16 was performed in a three-dimensionalrectangular chamber. The second reason is that in the grassseed experiment12 the ratio of the length to the thickness ofthe conductor was 7, but in the second experiment16 this ratio

Fig. 4. Equipotential lines dashed and electric field lines continuous in

the y0 plane. The bold horizontal lines represent the boundaries of the

strip at x/a1 and x/a 1. We assume A 2/A 13.6 and A 4/A30.

Fig. 5. Electric field lines of Fig. 4 overlaid on plate 6

of Ref. 13.

Fig. 6. Equipotential lines in the y0 plane overlaid on Fig. 3a of Ref. 16.

We use A 2/A 13.0 and A 4/A 30.


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was only 2.3, which means that boundary effects near z

andz are more important. These boundary effects were

not considered in our calculations.One of the main aspects of this work is that we succeeded

in finding a theoretical model yielding reasonable resultswhich were compared with two different experiments al-ready published in the literature. As discussed above, theseexperiments mapped the electric field lines and the equipo-tential lines inside and outside the regions of steady currentsin conductors. The geometry considered here had never beendealt with in this problem before. In order to obtain thisresult it was necessary to use elliptic-cylindrical coordinates

(, , z). The general solution for the potential in terms of

these variables is reasonably simple, namely, (A 1

A2)(A 3zA 4). When expressed in terms of the usual Car-

tesian coordinates (x , y , z) the solution takes the compli-

cated form of Eq. 5. We could not obtain this solutionworking only with cartesian coordinates. In this problem thepure cylindrical coordinates are not so practical as well. Thesituation described here shows an important example of theusefulness of the elliptic-cylindrical coordinates in dealingwith reasonably simple problems of physics.

ACKNOWLEDGMENTS

The authors thank the referees for important suggestions.JAH thanks CNPq Brasil for financial support. AKTAthanks the Alexander von Humboldt Foundation of Germanyfor a research fellowship during which this work was accom-plished.

aElectronic mail: [email protected] address: Instituto de Fsica Gleb Wataghin, Universidade

Estadual de Campinas Unicamp, 13083-970 Campinas, Sao Paulo, Brazil;

electronic mail: [email protected]

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Am-J-Phys-V71-p938-942(2003)