The Electric Field

If we haveseveralpoint chargesq₁,q₂, . . . ,qn, at distances

r

1,

r

2, . . . ,

r

^{n from}

Q, the total force onQis evidently F=F₁+F₂+. . .= 1

4π0

q1Q

r

²1

ˆr

1+q2Q

r

²2

ˆr

2+. . .

= Q 4π0

q1

r

²1

ˆr

^1+q

r

²2²

ˆr

^2+q

r

²3³

ˆr

^{3+. . .,}

or

F=QE, (2.3)

where

E(r)≡ 1 4π0

n i=1

qi

r

²i

ˆr

^{i. (2.4)}

Eis called theelectric fieldof the source charges. Notice that it is a function of position (r), because the separation vectors

r

ⁱ depend on the location of thefield point P (Fig. 2.3). But it makes no reference to the test charge Q. The electric field is a vector quantity that varies from point to point and is determined by the configuration of source charges; physically,E(r)is the force per unit charge that would be exerted on a test charge, if you were to place one atP.

What exactlyisan electric field? I have deliberately begun with what you might call the “minimal” interpretation ofE, as an intermediate step in the calculation of electric forces. But I encourage you to think of the field as a “real” physical

z

x y

P Source point q₁

r

i

r⬘i r

q₂ q_i Field

point

FIGURE 2.3

entity, filling the space around electric charges. Maxwell himself came to believe that electric and magnetic fields are stresses and strains in an invisible primordial jellylike “ether.” Special relativity has forced us to abandon the notion of ether, and with it Maxwell’s mechanical interpretation of electromagnetic fields. (It is even possible, though cumbersome, to formulate classical electrodynamics as an

“action-at-a-distance” theory, and dispense with the field concept altogether.) I can’t tell you, then, what a fieldis—only how to calculate it and what it can do for you once you’ve got it.

Example 2.1. Find the electric field a distancez above the midpoint between two equal charges (q), a distancedapart (Fig. 2.4a).

Solution

LetE₁be the field of the left charge alone, andE₂that of the right charge alone (Fig. 2.4b). Adding them (vectorially), the horizontal components cancel and the vertical components conspire:

Ez=2 1 4π0

q

r

^{2 cosθ.}

Here

r

=

z²+(d/2)²and cosθ=z/

r

^{, so}

E= 1 4π0

2q z

z²+(d/2)²_3/2z.ˆ

Check:Whenzd you’re so far away that it just looks like a single charge 2q, so the field should reduce toE=_4π¹₀^2q_z2 z. And itˆ does(just setd→0 in the formula).

q x z

q

P z

d/2 d/2 (a)

q x q

θ z

d/2 d/2 (b) E₂ E₁

E

r

FIGURE 2.4

Problem 2.2Find the electric field (magnitude and direction) a distancezabove the midpoint between equal and opposite charges (±q), a distancedapart (same as Example 2.1, except that the charge atx= +d/2 is−q).

r r

r

^P

P P dq

da⬘

dl⬘

dτ⬘

(a) Continuous distribution

(c) Surface charge, σ (d) Volume charge, ρ

r

^P

(b) Line charge, λ

FIGURE 2.5 2.1.4 Continuous Charge Distributions

Our definition of the electric field (Eq. 2.4) assumes that the source of the field is a collection of discrete point chargesqi. If, instead, the charge is distributed continuously over some region, the sum becomes an integral (Fig. 2.5a):

E(r)= 1 4π0

1

r

²

ˆr

^{dq. (2.5)}

If the charge is spread out along aline (Fig. 2.5b), with charge-per-unit-length λ, then dq = λdl (where dl is an element of length along the line); if the charge is smeared out over asurface(Fig. 2.5c), with charge-per-unit-areaσ, then dq=σda(wheredais an element of area on the surface); and if the charge fills avolume(Fig. 2.5d), with charge-per-unit-volumeρ, thendq=ρdτ(wheredτ is an element of volume):

dq→λdl∼σda∼ρdτ. Thus the electric field of a line charge is

E(r)= 1 4π0

λ(r)

r

²

ˆr

^{dl; (2.6)}

for a surface charge,

E(r)= 1 4π0

σ(r)

r

²

ˆr

^da; (2.7)

and for a volume charge,

E(r)= 1 4π0

ρ(r)

r

²

ˆr

^dτ. (2.8)

Equation 2.8 itself is often referred to as “Coulomb’s law,” because it is such a short step from the original (2.1), and because a volume charge is in a sense the most general and realistic case. Please note carefully the meaning of

r

^{in these}

formulas. Originally, in Eq. 2.4,

r

ⁱ stood for the vector from the source charge qi to the field pointr. Correspondingly, in Eqs. 2.5–2.8,

r

is the vector fromdq (therefore fromdl,da, ordτ) to the field pointr.²

Example 2.2. Find the electric field a distancezabove the midpoint of a straight line segment of length 2Lthat carries a uniform line chargeλ(Fig. 2.6).

x dx

r

P

x z

FIGURE 2.6 Solution

The simplest method is to chop the line into symmetrically placed pairs (at±x), quote the result of Ex. 2.1 (withd/2→x,q →λd x), and integrate (x:0→ L).

But here’s a more general approach:³

r=zzˆ, r=xxˆ, dl=d x;

r

^{=r−r=zzˆ−xx,ˆ}

r

^=z2+x2,

r

^ˆ=

r

r

^{= √zˆz}z²⁻+^xx^xˆ2. E= 1

4π0

L

−L

λ z²+x²

zzˆ−xxˆ

√z²+x²d x

= λ 4π0

zˆz L

−L

1

(z²+x²)^3/2d x−xˆ L

−L

x

(z²+x²)^3/2d x

= λ 4π0

zzˆ

x z²√

z²+x^{2 L}

−L

−xˆ

− 1

√z²+x^{2 L}

−L

= 1 4π0

2λL z√

z²+L² ˆz.

2Warning:The unit vector

ˆr

^isnotconstant; itsdirectiondepends on the source pointr, and hence itcannot be taken outside the integrals(Eqs. 2.5–2.8). In practice,you must work with Cartesian components(xˆ,yˆ,zˆareconstant, anddocome out), even if you use curvilinear coordinates to perform the integration.

3Ordinarily I’ll put a prime on the source coordinates, but where no confusion can arise I’ll remove the prime to simplify the notation.

For points far from the line(zL), E∼= 1

4π0

2λL z² .

This makes sense: From far away the line looks like a point chargeq =2λL. In the limit L → ∞, on the other hand, we obtain the field of an infinite straight wire:

E= 1 4π0

2λ

z . (2.9)

Problem 2.3Find the electric field a distancezabove one end of a straight line segment of lengthL(Fig. 2.7) that carries a uniform line chargeλ. Check that your formula is consistent with what you would expect for the casezL.

P z

L FIGURE 2.7

P z

a FIGURE 2.8

P z

r

FIGURE 2.9 Problem 2.4Find the electric field a distancezabove the center of a square loop (sidea) carrying uniform line chargeλ(Fig. 2.8). [Hint:Use the result of Ex. 2.2.]

Problem 2.5Find the electric field a distancezabove the center of a circular loop of radiusr(Fig. 2.9) that carries a uniform line chargeλ.

Problem 2.6Find the electric field a distancezabove the center of a flat circular disk of radiusR(Fig. 2.10) that carries a uniform surface chargeσ. What does your formula give in the limitR→ ∞? Also check the casezR.

Problem 2.7Find the electric field a distancezfrom the center of a spherical surface

!

of radiusR(Fig. 2.11) that carries a uniform charge densityσ. Treat the casez<R (inside) as well asz> R(outside). Express your answers in terms of the total charge qon the sphere. [Hint:Use the law of cosines to write

r

in terms of Randθ. Be sure to take thepositivesquare root:√

R²+z²−2Rz=(R−z)ifR>z, but it’s (z−R)ifR<z.]

Problem 2.8Use your result in Prob. 2.7 to find the field inside and outside a solid sphere of radius R that carries a uniform volume charge densityρ. Express your answers in terms of the total charge of the sphere, q. Draw a graph of |E|as a function of the distance from the center.

P z

R

FIGURE 2.10

r

x

y R

θ P

z

FIGURE 2.11

2.2 DIVERGENCE AND CURL OF ELECTROSTATIC FIELDS

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