68

Chapter 4

Sections 4-2: Charge and Current Distributions

CHAPTER 4

Problem 4.1 'A cube 2 m on a side is located in the first octant in a Cartesian

coor mate system, with one of its comers at the origin. Find the total chargecontained in the cube if the charge density is given by Pv = ;\y.e-2= (mC/m3).

Solution: For the cube shown in Fig. P4.1, application of Eq. (4.5) gives

Q= [ p"d'J/= j'2 12 [2 x:v-e-2=dxdydzJ'J/ x=O )=oJ==o2 12 12

(-1 2-3 -2=)= -Xye12

8= - (1 - e-4) -

x=O b=o 1==0 3 - 2.62 mC.

x

2m"."

".",---IIIII11m

o

- - - - - -~,,'" I" I

" I- - ( I

I II 12m

y

Figure P4.]: Cube of Problem 4.1.

Problem 4.2 Find the total charge contained in a cylindrical volume defined byr ::; 2 m and 0 ::; z ::; 3 m if Pv = 10rz (mC/m3).

Solution: For the cylinder shown in Fig. P4.2, application of Eq. (4.5) gives

[3 [21t 12Q= Jz=oJ$=O 1"=010rzrdrdq,dz

12 21t 3= (~r3q,~) = 2401t (mC) = 0.754 C.1"=0 $=0 ==0

CHAPTER 4

(b)

1rolL 1/"0lL (-PO )Q= pvd'V'= --2 21trdrdz

r=O ==0 r=O ==0 1 +r

Ioro r= -2npoL --2 dr = -1tpoLln(1 +r5)·o 1 +r

A upo 2J = Pvu = -z --2 (Aim),l+r

1= !J·ds

lro 121t(A upo ) A d dth= -z-- .zr r 'I'

1=0 $=0 I+r2

10/"0 r= -2nupo -1--2 dr = -nupo In(]+ r5) (A).o +r

(-PO )Pv = ] + r2

1 121t~It (A 25 ) A

1= J·ds= R- .(RR2sin8d8dcp)S $=0 9=0 R

It 121t

= - 25R<j>cos 81 I I = 500n = 1,570.8 (A).R=5 9=0 $=0

72

Problem 4.8 An electron beam shaped like a circular cylinder of radius 1() carries acharge density given by

where Po is a positive constant and the beam's axis is coincident with the z-axis.(a) Determine the total charge contained in length L ofthe beam.(b) Ifthe electrons are moving in the +z-direction with uniform speed ll, determine

the magnitude and direction of the current crossing the z-plane.

Solution:

(a)

00blem~ If J = R25 / R (A/m2), find I through the surface R = 5 m.Solution: Using Eq. (4.12), we have

Current direction is along -z.

73

m9- A square with sides 2 III each has a charge of20 pC at each of its fourDetermine the electric field at a point 5 III above the center of the square.

P(O,O,5)

J'

x

Figure P4.9: Square with charges at the comers.

The distance IRI between any of the charges and point P IS

y

Figure P4.1O: Locations of charges in Problem 4.10.

CHAPTER 4

x

Rl = -x 2 em

R2 = -y 2 em

RI Q-2cmQ

1 [3nC(-xo.02)] 3nC(-YO.02) A AE=41tE (0.02)3 + IAM\'I =-67.4(x+y) (kY/m)atR=O.

Employ Eq. (4.14) to find the force F = qE = - 202.2(x + y) (uN).

Problem 4.11 Charge q, = 4 ,uC is located at (I em, 1 cm,O) and charge CJ2

is located at (0,0,4 em). What should CJ2 be so that E at (0,2 cm,O) has noy-component?

Solution: For the configuration of Fig. P4.II, use of Eq. (4.19) gives

Problem 4.10 Three point charges, each with q = 3 nC, are located at the comerso a tnangle in the x-y plane, with one corner at the origin, another at (2 em, 0, 0),and the third at (0,2 cm,O). Find the force acting on the charge located at the origin.

Solution: Use Eg. (4.19) to determine the electric field at the origin due to the othertwo point charges [Fig. P4.1 0]:

IRI = VI2+12+52 = V27.

Q [Rl R2 R3 ~ ]E = 41t£o IRP + IRI3 + IRI3 + IRP

Q [-X-Y+Z5 x-y+z5 -x+y+z5 X+Y+Z5]= 41tEo (27)3/2 + (27)3/2 + (27)3/2 + (27)3/2A SQ ASx20,uC 0.71 -6 A

= Z. )3/2 = Z ()3/2 = - X 10 (V/m) = z25.61 (kY/m).27 1tto 27 1tEa 1tEo

74

CHAPTER 4

RI = -x + y(2-1) = (-x + y) em

R2 = (y2 - z4) em

y

x

Figure P4.1 I: Locations of charges in Problem 4.11.

75

~ 1 [4,uC(-i+Y)XIO-2 Q2(y2-z4) x 10-2]E(R = y2cm) = 41t€ (2 x 10-2)312 + (20 x 10-2)3/21

= -[-iI4.14 X 10-6 +Y(14. 14 x 10-6 + O.224q2)41t€ .

-z0.447q2] (Vim).

IfE.v = 0, then Q2 = -14.14 x 10-6 /0.224 ~ -63.13 (,uC).

~em 4.1V A line of charge with uniform density PI = 4 (,uClm) exists in airalong the z-axis between z = 0 and z = 5 em. Find E at (0, 10 em,O).

Solution: Use of Eq. (4.2 Ie) for the line of charge shown in Fig. P4.12 gives

1 t ~'PldJlE-- R--

- 41t€o I' R,2'

R' = YO.I-Zz

1 10.05 -6 (yO. 1 - Zz)=-4- (4xl0 )[( )2 _"".~dz1t€0 z=o 0.1 -t

[ ] 10.05

4 X 10-6 ylOz+z

- 41t€o vI(O.I)2 +z2 Iz=O

= 35.93 x 103 [Y4.47 - z 1.06] = Y160.7 x 103 - z38.1 X 103 (Vim).

76

x

5cm

d-o:.

IOcmy

CHAPTER 4

Figure P4.12: Line charge.

Problem 4.13 Electric charge is distributed along an arc located in the x-y planeand defined by r = 2 em and 0 :::;<jI:::;1t/4. If PI = 5 (pC/rn), find E at (O,O,z) andthen evaluate it at (a) the origin, (b) z = 5 em, and (c) z = -5 em.

Solution: For the arc of charge shown in Fig. P4.l3, dl = r d<jl= 0.02 d<jl, andR' = -iO.02eoslj>-YO.02sin<jl+iz. Use ofEq. (4.21 c) gives

1 1A/Pldl'E--- R--

- 41tEo II R/2

__ 1_ f1t/4 (-iO.02cos<jl-YO.02sin<jl+.zz) 002d- 41tEolcp=o PI "" ",,\'). ')\'1./') • <jI

898.8 [A A A] /= 11') -xO.014-yO.006+z0.78z (V m).((0.02)2 +z2)

(a) Atz= 0, E = -i1.6-YO.66 (MV/m).(b) Atz = 5 em, E = -i81.4 -Y33.7 +z226 (kV/m).(c) Atz = -5 em, E = -i81.4-Y33.7 -z226 (kV/m).

77

y1'2 em = I' 0.02 m

R' = -I' 0.02 + zz

x

Figure P4.13: Line charge along an arc.

L/2

eo = sin-I V,.2 + (L/2)2

Problem 4.14- A line of charge with uniform density PI extends between z = - L/2

and z = L 2 along the z-axis. Apply Coulomb's law to obtain an expression for theelectric field at any point P(,., </>, 0) on the x-y plane. Show that your result reduces tothe expression given by Eg. (4.33) as the length L is extended to infinity.

Solution: Consider an element of charge of height dz at height z. Call it element 1.The electric field at P due to this element is dEl. Similarly, an element at -zproduces dE2. These two electric fields have equal z-components, but in oppositedirections, and hence they will cancel. Their components along •.will add. Thus, thenet field due to both elements is

where the cos e factor provides the components of dEt and dE2 along r.Our integration variable is z, but it will be easier to integrate over the variable e

from e = 0 to

CHAPTER 4

LI2 ~ I,

x-y plane

CHAPTER 4

-L/2

Figure P4.14: Line charge of length L.

Li2

dz ITI"-1--T

Hence, with R = r/ cos e, and z = rtan e and dz = rse2 8d8, we have

lLI2 Ieeo lea PI cos3 eE =. dE = dE = r-2- --2-rsec2ede==0 6=0 0 TCfo r

PI 160= r-- cos8d82TCfor 0

A PI . 8 A PI LI2= r -- Sin 0 = r -- ----c====2TCfor 2TCfor Vr2 + (L/2)2

For L »r,

78

where Pso is a constant.

Solution: We start with the expression for dE given in Example 4-5 but we replacePs with PsoJ.2:

79

Fl = force on Q"

E =r~ (infinite line of charge).21t£o,.

If

4.16 Multiple charges at different locations are said to be in equilibriumthe force acting on anyone of them is identical in magnitude and direction to the

acting on any of the others. Suppose we have two negative charges, one locatedthe origin and carrying charge -ge, and the other located on the positive x-axis atdistance d from the first one and carrying charge -36e. Determine the location,

and magnitude of a third charge whose placement would bring the entireinto equilibrium.

CHAPTER 4

To perform the integration, we use

R2 = ,.2 +h2,

2RdR= 2rdr,

E = z Psoh r(.il+h2)1/2 (R2 - h2) dR2£0 lh R2

~ Psoh [1(.il+h2)1/2 1(.il+h2)1/2 h2 ]

=z- dR- -dR2£0 h h R2

= Z Psoh [ Va2 +h2 + 112 2h] .2£0 via2 + h2

Problem 4.15 Repeat Example 4-5 for the circular disk of charge of radius a, butin the present case assume the surface charge density to vary with,. as

Ps = PsoJ.2 (C/m2),

and

two right triangles are symmetrical and of equal corresponding sides, show that theelectric field is zero at the origin.

CHAPTER 4 81

x

PI

y

Figure P4.17: Kite-shaped arrangment of line charges for Problem 4.17.

'~Iution: The field due to an infinite line of charge is given by Eq. (4.33). In the··present case, the total E at the origin is

components of E, and Ez along x cancel and their components along -y add.EJ is along y because the line charge on the y-axis is negative. Hence,

A 2P/cos8 A 2pIE--y---+y--- 21tEoRt 21tEoR2

But cos 8 = R, / Rz. Hence,

Pmb1em ~ Three infinite lines of charge, PI, = 5 (nC/m), P/2 = -5 (nC/m), and(n 1m), are all parallel to the z-axis. If they pass through the respective points

Components of line charges ] and 3 along y cancel and components along x add.Hence, using Eq. (4.33),

CHAPTER 4

y

,O,b) ~El

P/2

E~P/x

Eo

J

'O.-b)~

Figure P4.18: Three parallel line charges.

Ph = 5 (nC/m),

Ph = -5 (nC/m),

Ph = Ph,

E = EI +E2 +E3.

82

(0, -b), (0,0), and (O,b) in thex-y plane, find the electric field at (a,O,O). Evaluateyour result for a = 2 cm and b = I em.

Solution:

E ~ 2Ph 9 +~ Ph=x---cos x----.21tEoR, 21tEoa

withcos9= va2a+b2 andR, =va2+lJ1,

E = -2:-~-[-a2-~-b-2 - ~] X 10-9 (Vim).

Figure P4.19: Horizontal strip of charge.

x

83

Ps

E = x2.70 (kV/m).

/////

For a = 2 cm and b = I cm,

CHAPTER 4

Solution: The strip of charge density Ps (C/m2) can be treated as a set of adjacent linecharges each of charge PI = Ps dy and width dy. At point P, the fields of line chargeat distance y and line charge at distance -y give contributions that cancel each otheralong y and add along z. For each such pair,

dE=i2Psdycos6.21ttoR

A horizontal strip lying in the x-y plane is of width d in theand infinitely long in the x-direction. If the strip is in air and has a

uniform charge distribution Ps, use Coulomb's law to obtain an explicit expressionfor the electric field at a point P located at a distance h above the centerline of thestrip. Extend your result to the special case where d is infinite and compare it with

.Eq. (4.25).