Problems
Application 2.1 Moving Point Charges
3.4 Gauss’ Law and Solid Angle
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68 ELECTROSTATICS: THE ELECTRIC FIELD PRODUCED BY STATIONARY CHARGE
the hole) with a surface charge density equal to the negative of surface charge density of the shell.
Becauseba, the spherical cap is essentially a disk of radiusb. Because|r−R| b, the electric field of the disk is essentially the same as the electric field of a point charge with the same total charge. From what we have said, the latter isQ= −π b2σ = −Qb2/4a2. Therefore, using a step function(r−a), which is equal to one ifr > aand is zero otherwise, the electric field observed atris
E(r)= Q 4π 0
ˆ r
r2(r−a)+ Q 4π 0
r−R
|r−R|3.
3.4 Gauss’ Law and Solid Angle 69
z r
(a)
(b)
(c)
Figure 3.8: Shaded areas are highly symmetrical distributions of charge: (a)ρ0(r, θ, φ)=ρ0(r);
(b)ρ0(ρ, φ, z)=ρ0(ρ); (c)ρ0(x, y, z)=ρ0(z). Dashed lines outline possible choices for Gaussian surfaces.
plus a divergent constant of integration as discussed for (3.37). Finally, an infinite slab of charge [Figure 3.8(c)] is invariant to translations along thex- andy-axes. Ifσ(z) is the charge per unit area enclosed by a parallelepiped of width 2z, Gauss’ law tells us that
E(r)=E(z)ˆz ⇒ E(z)=σ(z) 2ε0
z
|z|. (3.44)
Up to another (infinite) constant, the potential forσ(z)=σis ϕ(z)= − σ
2ε0
|z|. (3.45)
It is important to remark that the symmetry of a charge distribution, i.e., its invariance with respect to spatial operations like rotation, translation, and reflection, tells us only that these symmetry operations transform one solution forE(r) into another. The Gauss’ law solution isthesolution because it is unique by Helmholtz’ theorem.8
The potentials (3.41), (3.43), and (3.45) are also worthy of comment. The first of these—the point charge potential—is the building block we used to derive the superposition integral appropriate to situations where the source charge density possesses no particular symmetry:
ϕ(r)= 1 4π 0
d3r ρ(r)
|r−r|. (3.46)
Two-dimensional charge densities are invariant to translations along a fixed direction, say, ˆz. To find the corresponding potential, we can superpose line charge potentials like (3.43), each weighted by its own charge per unit length(x, y)dxdy. This produces the potential
ϕ(x, y)= − 1 2π 0
dx
dyln
(x−x)2+(y−y)2(x, y). (3.47) The charge density shown in Figure 3.8(b) is a special case of this situation where (x, y)= (
x2+y2). A similar argument applies to one-dimensional charge distributions that are invari- ant to translations alongx andy as in Figure 3.8(c). Superposing sheet potentials like (3.45) with charge per unit areaγ(z)dzgives the potential
ϕ(z)= − 1 20
dz|z−z|γ(z). (3.48)
8 The Helmholtz theorem (Section 1.9) can fail if the charge density does not go to zero fast enough at infinity. We can establish uniqueness for infinite lines and planes of charge as the limit of a sequence of charged lines and planes with finite extent.
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70 ELECTROSTATICS: THE ELECTRIC FIELD PRODUCED BY STATIONARY CHARGE
Example 3.4 Begin with the electric field and derive an expression for the potentialϕ(r) produced by a spherically symmetric charge distributionρ(r).
Solution: Let
Q(r)=4π r
0
ds s2ρ(s)
be the charge contained within a ball of radiusrcentered at the origin. Equation (3.40) gives the electric field due toQ(r) as
E(r)= Q(r) 4π 0r2r.ˆ Using (3.18) withd=drrˆandr0at infinity, the potential atris
ϕ(r)= − r
∞
d·E(r)= −1 0
r
∞
dr r2
r
0
ds s2ρ(s)= r
∞
d 1
r
u(r) where
u(r)= 1 0
r
0
ds s2ρ(s).
Integrating by parts gives
ϕ(r)= u(r) r
r
∞− r
∞
dr r
du dr or
ϕ(r)= 1 0r
r
0
drr2ρ(r)+ 1 0
∞
r
drrρ(r).
3.4.2 Matching Conditions for E(r)
We can use the field (3.44) to re-derive the matching conditions (2.49) for the electric field near an arbitrarily shaped surfaceSthat carries a surface charge densityσ(rS) (Figure 3.9). The key is to let E1andE2be the electric fields at points that are infinitesimally close to each other, but on opposite sides ofS.
E1andE2may each be decomposed into two contributions: (1) the fieldEdiskproduced by the tiny, flat disk of charge (centered onrS) shown shaded in Figure 3.9; and (2) the fieldESproduced by the surface Swith the tiny disk atrSremoved. When viewed at very close range,Ediskis indistinguishable from the field (3.44) produced by an infinite, flat sheet of charge with uniform charge densityσ=σ(rS).This field changes sign when the observation point passes through the disk. Conversely,ES is smooth and continuous when the observation point passes through the hole created by the removed disk. Hence, by superposition,
E1=ES− σ 20
ˆ
n1 and E2=ES− σ 20
ˆ
n2. (3.49)
3.4 Gauss’ Law and Solid Angle 71
2( ) E r
ˆ1 n
ˆ2 n
( )rS 1( ) E r
Figure 3.9: A surfaceSendowed with charge densityσ(rS). The unit normal vector ˆn1points outward from region 1. The unit normal vector ˆn2points outward from region 2.
Subtracting the two equations in (3.49) givesE1−E2=nˆ2σ/0. Taking the dot product and cross product of this expression with ˆn2produces the expected matching conditions,
ˆ
n2·[E1−E2]=σ(rS)/0 (3.50)
and
ˆ
n2×[E1−E2]=0. (3.51)
3.4.3 The Force on a Charged Surface
It is not immediately obvious how to calculate the Coulomb force on a charged surfaceS because (3.50) says that ˆn·Eis discontinuous there. On the other hand, no surface element can exert a force on itself. Therefore, the quantityESin (3.49) must be solely responsible for the force per unit area felt atrS. Since ˆn1= −nˆ2, the sum of the two equations in (3.49) givesES as the simple average ofE1 andE2. Therefore, the desired force density is
f=σES= 12σ(E1+E2). (3.52)
3.4.4 Solid Angle
This section introduces the geometrical concept ofsolid angleand uses it (together with Coulomb’s law for a point charge) to derive the integral form of Gauss’ law (3.39). The solid angle will return later when we discuss magnetostatics.
Figure 3.10 shows an origin of coordinatesOand vectorsrandrSthat label an observation point O and a point on a surfaceS, respectively. A set of rays fan out radially fromOto every point of S. By definition, thesolid angleS subtended bySatOis that part of the area of a sphere of unit radius centered atO that is cut off by the rays. In other words, we project every surface element dS=dSnˆonto the direction ofrS−r, divide by the square of this distance (because we are interested in projection onto a unit sphere), and sum over all such elements. This gives
S(r)=
S
dS(r)=
S
dS· rS−r
|rS−r|3. (3.53)
It is worth noting thatS(r)changes signwhen the observation pointrpasses through the surface S. This happens becauserS−rchanges direction (relative toS) whiledSdoes not. The direction of the latter is fixed (by convention) so that (3.53) is positive when the curvature ofStends to enclose
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S
S nˆ rS r
unit sphere O
r
rS
' O
| SS | r r r r
Figure 3.10: Sis the solid angle subtended at the observation pointOby the surfaceS. See text for discussion.
O
O S S
V V
Figure 3.11: A closed surfaceSthat encloses a volumeV. The solid angleS=0 if the observation pointOis outsideV.S=4πifOis insideV.
the observation point. We note in passing that ifOcoincides withO(sor=0) andSis a spherical surface centered atO(so ˆn=ˆrS) we get the familiar result
dS =dSnˆ·rˆS
rS2 = dS
rS2 =sinθ dθ dφ. (3.54) We now putO=Oand specialize to aclosedsurfaceSthat encloses a volumeV. The left side of Figure 3.11 shows that every ray fromOintersectsSan even number of times ifOis outside ofV. Adjacent intersections where the ray enters and exitsV make the same projection onto a unit sphere centered atObut cancel one another in the sum (3.53) because ˆn·rˆS changes sign. Conversely, ifO is inside ofV (right side of Figure 3.11), every ray makes an odd number of intersections and the set of intersections closest toOproject onto the complete unit sphere with surface area 4π. With a slight change in notation, this determines our final geometrical result:
S =
S
dS· ˆr r2 =
4π 0
O∈V
O /∈V . (3.55)
In light of Coulomb’s formula for the electric field of a point charge atO, (3.55) is exactly Gauss’ law for a point charge:
E(S)=
S
dS· E= q 4π ε0
S. (3.56)
3.4 Gauss’ Law and Solid Angle 73
The general result (3.39) follows by superposition. The reader should note the crucial role played in this demonstration by the inverse-square nature of the electric force. Gauss’ law is not valid for other laws of force.