Charged line segments and ellipsoidal equipotentials

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by  T L Curtright

Charged line segments and ellipsoidal equipotentials
T L Curtright§, N M Aden, X Chen, M J Haddad, S Karayev, D B Khadka, and J Li
Department of Physics, University of Miami
Coral Gables, FL 33124-8046, USA
§
6 [email protected]
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0 January 18, 2016
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n
a Abstract
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Thisisasurveyoftheelectrostaticpotentialsproducedbychargedstraight-linesegments,invarious
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numbers of spatial dimensions, with comparisons between uniformly charged segments and those hav-
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ing non-uniform linear charge distributions that give rise to ellipsoidal equipotentials surrounding the

] segments. A uniform linear distribution of charge is compatible with ellipsoidal equipotentials only for
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threedimensions. Inhigherdimensions,thelinearchargedensitygivingrisetoellipsoidalequipotentials
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iscounter-intuitive—thechargedistributionhasamaximumatthecenterofthesegmentandvanishes
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s attheendsofthesegment. Onlyintwodimensionsisthecontinuouschargedistributionintuitive—for
s thatonecaseofellipsoidalequipotentials,thechargeispeakedattheendsofthesegmentandminimized
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at the center.
l
c
.
s
c Contents
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y 1 Introduction 1
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p 2 Electrostatics in two dimensions 2
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1 3 Point charge in D dimensions 13
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7 4 Uniformly charged line segments for all dimensions D >2 14
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0 5 Ellipsoidal equipotentials for all dimensions D >2 19
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6 Generalizations 25
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0 7 Summary 26
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:
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1 Introduction
Ithasbecomeawidespreadpracticetostudythephysicsofsystemsinvariousnumbersofspatialdimensions,
not necessarily D =3. For instance, graphene with D =2, and string or membrane theory with D as high
as25, aretwoexamplesthatimmediatelycometomindforboththeirexperimentalandtheoreticalinterest.
Moreover, geometric ideas provide a common framework used to pursue such studies. A pedagogical goal
of this paper is to encourage students to think along these lines in the context of a familiar subject —
electrostatics.
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Forexample,inthreespatialdimensionsauniformlychargedstraight-linesegmentgivesrisetoanelectric
potentialΦwhoseequipotentialsurfacesareprolateellipsoidsofrevolutionaboutthesegment,withtheends
of the segment providing the foci of the ellipsoid. As an immediate consequence of the geometry for these
→− →−
prolate ellipsoidal equipotentials, the associated electric field E = −∇Φ — always normal to surfaces of
constant Φ — has at any observation point a direction that bisects the angle formed by the pair of lines
from the observation point to each of the two foci of the ellipsoid. This beautiful electrostatic example was
presentedbyGeorgeGreenin1828[1], andithasbeendiscussedinmanybookssincethen[2]-[10]including
at least two texts from this century [11]. While the straight-line segment is an idealization, nevertheless
it provides insight into the behavior of real thin-wire conductors, especially upon approximating those real
wires as very narrow, “needle-like” ellipsoids.
Butasitturnsout,thelinesegmentprobleminthreedimensionsisaveryspecialcase,insomesensethe
most ideal of all possible electrostatic worlds. In any other dimension of space, uniformly charged segments
do not produce ellipsoidal equipotentials. Conversely, in any other dimension, if the equipotentials are
ellipsoidalaboutalinearlydistributedstraight-linesegmentofcharge,thenthatchargedistributioncannot
be uniform.
One need look no farther than two dimensional systems to see clearly that there are differences between
uniformly charged segments and those with charge distributed so as to produce ellipsoidal equipotentials.
Indeed, D = 2 is the only ellipsoidal equipotential case which is intuitive in the sense that the associated
linear charge distribution has maxima at the ends of the segment, as one might naively expect from the
repulsive force between like charges placed on a segment of a real, thin conductor at a finite potential. In
contrast,thedistributionofchargeneededtoproduceellipsoidalequipotentialsiscounter-intuitiveinhigher
dimensions. ToproducesuchequipotentialsforD >3thechargedistributionmusthaveamaximumatthe
center and vanish at the ends of the segment.
We begin our discussion in Section 2 in two dimensions, where the two types of charged segments are
readily analyzed. Then we compare and contrast uniformly charged line segments with those admitting
ellipsoidal equipotentials for any number of spatial dimensions. As a preliminary, we first discuss briefly in
Section 3 the potential of a point charge in D dimensions. We then use this information in Section 4 to
compute the potentials and electric fields for uniformly charged line segments. We continue in Section 5 by
considering systems with ellipsoidal equipotentials in D dimensions. We then determine the linear charge
distributions that produce such potential configurations, and we find the remarkable result that the linear
charge density giving rise to ellipsoidal equipotentials is peaked at the center of the segment, for D >3.
The discussion of general D affords the opportunity to illustrate how continuous D can be used as a
mathematical device to regulate singular behavior. This too is a widespread practice in theoretical physics.
We use D in this way in Section 5 of the paper to interpolate continuously between intuitive and counter-
intuitive charge distributions for segments with ellipsoidal equipotentials.
Finally, inSection6, weinvokewell-knownmethodstoindicatehowthevariouslinesegmentresultsalso
give solutions to a class of electrostatic boundary value problems where the charge is moved outward and
distributed on one of the equipotential surfaces that surrounds the original segment.
2 Electrostatics in two dimensions
The point-particle electric potential in 2D is well-known to be logarithmic. For a point charge Q located at
the origin, up to a constant R that sets the distance scale,
Φ (→−r)=kQln(R/r) , ∇2Φ (→−r)=−2πkQ δ2(→−r) , (1)
point point
where k is the two dimensional analogue of Coulomb’s constant.
By linear superposition, a finite length, uniformly charged line segment, with constant charge/length λ
distributed on the x axis for −L/2≤x≤L/2, produces a potential
(cid:18) (cid:112) u(cid:19)(cid:12)(cid:12)u=−x+L/2
Φline(x,y)=−kλ uln u2+y2−u+yarctan y (cid:12)(cid:12) +kλLln(R) . (2)
u=−x−L/2
2
This result may be established by integrating the contributions of infinitesimal point-like bits of charge that
make up the segment, using (1) and the indefinite integral
(cid:90) (cid:16)(cid:112) (cid:17) (cid:16)(cid:112) (cid:17) u
ln u2+y2 du=uln u2+y2 −u+yarctan . (3)
y
Written out in full,
(cid:18) (cid:18)x− 1L(cid:19) (cid:18)x+ 1L(cid:19)(cid:19)
Φ (x,y)=kλ L+yarctan 2 −yarctan 2 (4)
line y y
 (cid:115)   (cid:115) 
(cid:18) 1 (cid:19) 1 (cid:18) 1 (cid:19)2 (cid:18) 1 (cid:19) 1 (cid:18) 1 (cid:19)2
+ x− L kλln x− L +y2− x+ L kλln x+ L +y2 .
2 R 2 2 R 2
A plot of the potential surface shows the essential features.
x y
Figure 1: Potential surface for a uniformly charged line segment in 2D.
The top of the potential surface is curved and not a straight line, indicating that the charged segment
itself is not an equipotential. This follows analytically from (4). Although points on the segment are not
at the same potential, they are all at finite values of the potential, for D =2. Explicitly, for y =0 and all
x,
Φline(x,0)=kλL−kλ(cid:0)21L+x(cid:1)ln(cid:0)(cid:12)(cid:12)12L+x(cid:12)(cid:12)/R(cid:1)−kλ(cid:0)12L−x(cid:1)ln(cid:0)(cid:12)(cid:12)12L−x(cid:12)(cid:12)/R(cid:1) . (5)
We plot f(x)=Φ (x,0)/kλ versus x to show the shape of the potential along the x-axis, for L=2 and
line
R=1.
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f
2
4 2 2 4
x
2
Figure 2: Uniformly charged line segment potential for D =2, along the x-axis.
Ontheotherhand,transversetothex-axisthepotentialhasadiscontinuousslopeacrossthelinesegment.
That is to say, the electric field normal to the line segment, E , is discontinuous due to the presence of the
y
charge density on the segment. For example, for x=0 this transverse profile is given by
(cid:20) (cid:18)(cid:113) (cid:19) (cid:21)
Φ(0,y)=kλ L−Lln 1L2+y2/R −2yarctan(cid:0)1L/y(cid:1) . (6)
4 2
We also plot g(y) = Φ (0,y)/kλ versus y to show the shape of the potential along the y-axis, for L = 2
line
and R=1.
g
2
4 2 2 4
y
2
Figure 3: Uniformly charged line segment potential for D =2, along the y-axis.
Moreover, the equipotentials are not ellipses surrounding the segment on the xy-plane, as is especially
clear for points close to the segment. The potential contours actually intersect the segment. A view of the
potentialcontoursfrombelow thepotentialsurfaceshowsthesefeaturesgraphically. (Seetheplottofollow.
But note the view in that plot is actually an orthogonal projection of the contours onto the xy-plane, and
not the true perspective of an observer on the potential axis a finite distance below that plane.)
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x
2 1 0 1 2
y
Figure 4: Potential contours for a uniformly charged line segment in 2D.
The components of the electric field produced by the segment are given by
(cid:118) 
∂ (cid:117)(cid:117)(cid:0)x+ 1L(cid:1)2+y2
Ex(x,y)=−∂xΦline(x,y)=kλln(cid:116)(cid:0)x− 21L(cid:1)2+y2 , (7)
2
(cid:18) (cid:19) (cid:18) (cid:19)
∂ L+2x L−2x
E (x,y)=− Φ (x,y)=kλarctan +kλarctan , (8)
y ∂y line 2y 2y
→−
where we have used d arctanu = 1 . Note that E (x,y) is independent of the scale R, since changing
du u2+1
→−
R just amounts to adding a constant to the potential. But of course E (x,y) does depend on L as this sets
the physical length scale for the system.
(cid:82) →− →−
As a check, in the units we have chosen the charge density along the line follows from ∇ · E dxdy =
(cid:82)
2πk ρ dxdy, integrated over a horizontal rectangle containing an infinitesimal portion of the x-axis, as the
height of the rectangle is taken to zero. Thus, using lim arctanz =±π/2,
z→±∞

0 if x>L/2

2πk λ(x)=2limE (x,y)=2kλ× π if −L/2<x<L/2 . (9)
y
y→0  0 if x<−L/2
That is to say, the charge/length between −L/2 and L/2 is the constant λ, as expected. Elsewhere, the
→−
charge density vanishes, as follows from ∇2Φ (r)=0 for all points not coincident with the segment.
line
It is instructive to make a vector field plot of (E (x,y),E (x,y)), especially near the charged segment.
x y
Again it is evident graphically that the segment itself is not an equipotential, since the electric field lines
are not perpendicular to the segment as the x-axis is approached for −L/2<x<L/2, except at the single
point x=0. Here are vector field plots for L=2.
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2
y
1
2 1 1 2
x
1
2
y 0.1
1.2 1.0 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0 1.2
x
0.1
→−
Figures 5 & 6: Vector field plots of E for a uniformly charged line segment situated between x=−1 and
x=+1. The field is evaluated at the center of each arrow.
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In contrast to the uniformly charged segment in 2D, consider a distribution of charge along the segment
such that equipotentials are ellipsoidal. The relevant charge distribution turns out to be

0 if x>L/2
2λL 1 
λ(x)= √ 1 if −L/2<x<L/2 , (10)
π L2−4x2  0 if x<−L/2
as we shall confirm in the following. Since (cid:82)1 √ 1 ds = π, the total charge on the segment is still
−1 1−s2
(cid:82)L/2
Q= λ(x) dx=λL, the same as for the uniformly charged case.
−L/2
The corresponding potential is now
(cid:18) (cid:19)
4R
Φ (x,y)=kλLln √ , (11)
line
s+ s2−L2
where s is a sum of two distances, from the observation point (x,y) to each of the two ends of the segment.
That is,
(cid:115)
(cid:18) 1 (cid:19)2
s=r +r , r = x± L +y2 . (12)
− + ± 2
Because the positional dependence of the potential is given entirely by s, the equipotentials are ellipses on
the xy-plane, with the ends of the segment serving as the foci of each equipotential ellipse. Again, we plot
the potential surface.
x y
Figure 7: Potential surface for a non-uniformly charged line segment in 2D, with λ(x)= √2λL/π .
L2−4x2
The top of the potential surface is now a straight line, indicating that the charged segment itself is an
equipotential,andthepointsonthesegmentareatafinite potentialforD =2,namely,Φ=kλLln(4R/L).
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This follows analytically from (11). Explicitly, for y =0 and all x,
 
4R
Φline(x,0)=kλLln(cid:12)(cid:12)x− 1 L(cid:12)(cid:12)+(cid:12)(cid:12)x+ 1 L(cid:12)(cid:12)+(cid:113)2(cid:0)x2− 1L2+(cid:12)(cid:12)x2− 1 L2(cid:12)(cid:12)(cid:1) . (13)
2 2 4 4
We plot f(x) = 1 Φ (x,0) versus x to show the shape of the potential along the x-axis, for L = 2 and
kλ line
R=1.
f
1
4 2 2 4
x
1
2
3
Figure 8: Non-uniformly charged line segment potential for D =2, along the x-axis.
Transverse to the x-axis, the potential again has a discontinuous slope across the line segment due to the
presence of the charge density on the segment. For example, for x=0 this transverse profile is given by
(cid:32) (cid:33)
4R
Φ (0,y)=Lkλln . (14)
line (cid:112)
L2+4y2+2|y|
We plot g(y) = 1 Φ (0,y) versus y to show the shape of the potential along the y-axis, for L = 2 and
kλ line
R=1.
g
1
4 2 2 4
y
1
2
3
Figure 9: Non-uniformly charged line segment potential for D =2, along the y-axis.
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A view of the potential contours from below the potential surface shows the equipotential ellipses. (But
again note the view in that plot is actually an orthogonal projection of the contours onto the xy-plane, and
not the true perspective of an observer on the potential axis a finite distance below that plane.)
x
2 1 0 1 2
y
Figure 10: Ellipsoidal potential contours for a non-uniformly charged line segment in 2D.
The components of the electric field are now
∂ kλL ∂s
E (x,y)=− Φ (x,y)= √ , (15)
x ∂x line s2−L2∂x
∂ kλL ∂s
E (x,y)=− Φ (x,y)= √ , (16)
y ∂y line s2−L2∂y
whose final form in rectangular coordinates follows from
∂s x− 1 L x+ 1 L 1 (cid:18) 1 (cid:19) ∂s y y 1
= 2 + 2 = sx− (r −r )L , = + = sy . (17)
∂x r r r r 2 + − ∂y r r r r
− + + − − + + −
In terms of x, y, r , and r ,
+ −
(cid:18)(cid:18) (cid:19) (cid:18) (cid:19) (cid:19)
kλL 1 1
E (x,y)= x− L r + x+ L r , (18)
x r r (cid:113)(r +r )2−L2 2 + 2 −
+ − − +
kλL
E (x,y)= (r +r )y . (19)
y (cid:113) − +
r r (r +r )2−L2
+ − − +
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Writing everything out in terms of x and y gives
(cid:115)  (cid:115) 
(cid:18)1 (cid:19)2 1 (cid:18)1 (cid:19)2
r+r− = 4 L2+x2+y2 −L2x2, (r−+r+)2−L2 =2x2+y2− 4L2+ 4 L2+x2+y2 −L2x2 .
(20)
→−
The components of E are not elegant in these coordinates [13] but their properties are fully encoded and
amenable to machine computation.
(cid:113) (cid:113)
kλL (cid:0)x+ 1 L(cid:1) (cid:0)x− 1 L(cid:1)2+y2+(cid:0)x− 1 L(cid:1) (cid:0)x+ 1 L(cid:1)2+y2
E (x,y)= √ 2 2 2 2 , (21)
x (cid:114)
2 (cid:113) (cid:113)
(cid:0)1 L2+x2+y2(cid:1)2−L2x2 x2+y2− 1L2+ (cid:0)1 L2+x2+y2(cid:1)2−L2x2
4 4 4
(cid:18)(cid:113) (cid:113) (cid:19)
(cid:0)x− 1 L(cid:1)2+y2+ (cid:0)x+ 1 L(cid:1)2+y2 y
kλL 2 2
E (x,y)= √ . (22)
y (cid:114)
2 (cid:113) (cid:113)
(cid:0)1 L2+x2+y2(cid:1)2−L2x2 x2+y2− 1L2+ (cid:0)1 L2+x2+y2(cid:1)2−L2x2
4 4 4
Although the geometric features of the electric field may not be transparent from these expressions, never-
→−
theless the fact that the equipotentials are ellipsoidal allows one to immediately visualize the direction of E
as normal to those surfaces of constant Φ.
(cid:82) →− →− (cid:82)
To confirm the charge density along the line, we again use ∇ · E dxdy =2πk ρ dxdy, and integrate
over a horizontal rectangle containing an infinitesimal portion of the x-axis, as the height of the rectangle is
taken to zero. The crucial features here are s−L → |2x−L| > 0 as y → 0 for |x| > L/2 (away from the
segment), but s−L→0 as y →0 for −L/2<x<L/2 (on the segment). More precisely, as points on the
segment are approached transversely,
(cid:18)1 (cid:19)(cid:115) y2 (cid:18)1 (cid:19)(cid:115) y2
s= L−x 1+ + L+x 1+
2 (cid:0)1 L−x(cid:1)2 2 (cid:0)1 L+x(cid:1)2
2 2
(cid:18) (cid:19) (cid:18) (cid:19)
∼ L+ 1 1 + 1 y2+O(cid:0)y4(cid:1)=L+ 1 L 1 y2+O(cid:0)y4(cid:1) . (23)
y→0 2 1 L−x 1 L+x 2 1 L2−x2
2 2 4
Thus for −L/2<x<L/2,
(cid:18) (cid:19) (cid:114)
y 1 y 1 y 1 1
lim √ = √ lim √ = √ lim = L2−x2 . (24)
(cid:114)
y→0 s2−L2 2Ly→0 s−L 2Ly→0 (cid:16) (cid:17) L 4
1 L 1 y2
2 1 L2−x2
4
Combining this with the more obvious lim (r r )= 1 L2−x2 for −L/2<x<L/2 gives
y→0 + − 4

(cid:18)2kλLs y (cid:19) 2kλL  0 if x>L/2
2πk λ(x)=2limE (x,y)= lim √ = × 1 if −L/2<x<L/2 .
y→0 y y→0 r+r− s2−L2 (cid:113)1 L2−x2  0 if x<−L/2
4
(25)
That is to say, the charge/length between −L/2 and L/2 is
2λL
λ(x)= √ , (26)
π L2−4x2
→−
as anticipated above in (10). Elsewhere, the charge density vanishes, again as follows from ∇2Φ (r)=0
line
for all points not coincident with the segment.
Oncemoreitisinstructivetomakeavectorfieldplotof(E (x,y),E (x,y)),especiallynearthecharged
x y
segment. The fact that the segment itself is an equipotential is evident graphically since the electric field
lines are perpendicular to the segment as the x-axis is approached for −L/2 < x < L/2. Here are vector
field plots for L=2.
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