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

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

1312ft12Q= lOrzrdrd¢Jdzz=o 41=0 r=O

/' ~ r= (~r'$z') '=0 .~ ,~o= 2401t (mC) =0.754 c.

CHAPTER 4

y

- - - - - -.,,," I.- I

.- ,- - ( ,I ,, ,2m

2m

z

.­.-

.-,,---,,,I /0I

2m'

Figure P4.1: Cube of Problem 4.1.

x

144

Chapter 4

Sections 4-2: Charge and Current Distributions

~blem 9A cube 2 m on a side is located in the first octant in a Cartesiancoordinate system, with one of its corners at the origin. Find the total chargecontained in the cube if the charge density is given by pv =xTe-2z (mC/m3).

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

Q= r Pvd'll= r2 r2 r2 xy'le-2ZdxdydzJ'lI Jx=oJy=oJz=o

12 2 12= ( 1; XZ1e-2z) = ~(1 - e-4) = 2.62me.=0 y=0 ;:=0

149

4·3: Coulomb's Law

A square with sides 2 m each has a charge of 20 pC at each of its fourthe electric field at a point 5 m above the center of the square.

z

P(0,Q,5)

y

x

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

distance JR/ between any of the charges and point P IS

y

CHAP1BR 4

x

RI==-x2cmA .

R2 == -y 2 cm

RI Q­2cmQ

150

~blem 449/ Three point charges, each with q = 3 nC. are located at the cornersof a triangle in the x-y plane. with one corner at the origin. another at (2 cm,O,O),and the third at (0,2 cm,O). Find the force acting on the charge located at the origin.

Solution: Use Eq. (4.19) to determine the electric field at the origin due to the othertwo point charges [Fig. P4.10):

1 [3nC (-XO.02)] 3nC (-YO.02)E= 41tE (0.02p + IAM\1 =-67.4(x+y) (kV/m)atR=O.

Employ Eq. (4.14) to find the force F = qE = -2022(x+y) (uN).

Figure P4.1 0: Locations of charges in Problem 4.10.

IRI = \1'12 + }2 +52 =..tfi.

Q r RI R2 R3 ~ ]E = 41t£o L IRI3+ IRp + IRp + IRp

Q [-X-Y+Z5 x-y+Z5 -x+y+z5 X+Y+ZS]= 41"C€{) (27)3/2 + (27)3/2 + (27)3/2 + (27)3/2A 5Q ~ 5x20,uC 0.71 -6 A

= z. )3/2 = Z (P/2 = - X 10 (VIm) = z25.61 (kV/m).27 1reQ 27 1tC:o 1&£0

Problem 4.11 Charge ql = 4 ,uC is located at (1 cm, 1 em,O) and charge q2

is located at (0,0,4 ern). What should q2 be so that E at (0,2 cm,O) has noy-component?

Solution: For the configuration of Fig. P4.11, use ofEq. (4.19) gives

Cfroblem ~ A line of charge with unifonn density PI = 4 (pC/m) exists in airalong the z-axis between z = 0 and z = 5 em. Find E at (0,10 em,O).

Solution: Use ofEq. (4.21c) for the line of eharge shown in Fig. P4.12 gives

E= _1_ f A' PI dlf41tC{} III Rf2'

R' = YO.l-iz1 ,0.05 I A • _ • 0-6\ (yo.] - zz) J

= 41t€o Jz=o t'f X lU -} {(o.IF +zz]3/z at

= 4 X 10-6 [ ylOz+z ] r"41tC{) y'(O.I)2+zz z=o

= 35.93 X 103 [y4,47 - z1.06] = y 160.7 X 103 -z38.1 X 103 (VIm).

15]

y

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

Rz = (12 - z4) em

Figure P4.11: Locations of charges in Problem 4.11.

z

4 em t Ql

.x

CHAPTER 4

E(R= A2cm) = _1_ [4,uC(-X+Y) x 10-2 qz(y2-Z4) x 10-2]y 4ne (2 x 10-2)3/2 + (20 x lO-z)3/21

= 4ne[-xI4.14 x 10-6 +Y(14.14 x 10-6 +0.224qz)

- ZO.447qz] (VIm).

If Ey =0, then q2 = -14.14 X 10-6 /0.224 ~ -63.13 (PC).

CHAPTER 4

y

Figure P4.12: Line charge.

Scm

x

E- _1_ {ft/Pidl'- 4rc£0Jl' R,2

1 l1t/4 (-xO.02eos<j>-YO.02sin<j>+zz)= 41t" PI u~ ~~,., . -""I., O.02d<\>cO $=0

898.8 [A 0 AO 06 AO ]= ~/.,-xO. 14-y.0 +z .78z (Vim).((0.02)2 +zzr

....(a) At z = 0, E = -x 1.6 - 5'0.66 (MV/m).(b) At z = 5 em, E = -x81.4- 5'33.7 +i226 (kV/m).(c) At z = -5 em, E = -x81.4 - Y33.7 - i226 (kV/m).

152

Problem 4.13 Electric charge is distributed along an arc located in the x-y planeand defined by r = 2 em and 0:::; <1>:::;rc/4. If Pi = 5 CuC/m), find Eat (O,O,z) andthen evaluate it at (a) the origin, (b) z = Sem, and (c) z = -5 em.

Solution: For the arc of charge shown in Fig. P4.13, dl = r d<jJ= 0.02 d<l>, andR' = -xO.02cos<j> - 5'0.02 sin <jJ+ Zz. Use ofEq. (4.21c) gives

157

x

Pl

y

Pi

Three infinite lines of charge, PI, =5 (nC/m), Ph = -5 (nClm). andare all parallel to the z-axis. If they pass through the respective points

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

CHAPTER 4

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

Solution": The field due to an infinite line of chmge is given by Eq. (4.33). In thepresent case, the total E at the origin is

The components of El and E2 along x cancel and their components along -y add.Also. E3 is along y because the line chatge on the y-axis is negative. Hence.

"""But cos e =RJ/ R2. Hence.

E=_y~Rl+y~=O.'ft€oR1 R2 1f£oR2

(0,-b), (0,0), and (0, b) in the x-y plane, find the electric field at (a,0, 0). Evaluateyour result for a = 2 cm and b = 1 em.

Solution:

CHAPTER 4

y

(O,b)KE!

P12

E~p/x

E3(O'.b)~

Figure P4.18: Three parallel line charges.

158

PI, = 5 (nC/m),

Pi2 = -5 (nC/m),

Pi3 = Pi),

E=EI +E2+E3.

Components of line charges 1 and 3 along y cancel and components along x add.Hence, using Eq. (4.33),

E A 2p/, e A Ph=x2 R cos +x-2--·mo I moa

with cose = a and RI = va2 +P,..;a2 +b2

I? _ i5 r 2a~ ~ 21t€o la2 +b2

z

x

159

Ps

E = i2.70 (kV/m).

//I//

Figure P4.] 9: Horizontal strip of charge.

The strip of charge density Ps (C/m2) can be treated as a set of adjacent lineeach of charge PI =Psdy and width dy. At point P, the fields of line chatge

y and line charge at distance -y give contributions that cancel each othery. and add along Z. For each such pair.

dE =i 2PsdycosB .21re{)R

CHAPTER 4

For a = 2 cm and b = I em.

Problem 4.19 A horizontal strip lying in the x-y plane is of width d in they-direction and infinitely long in the x-direction. If the strip is in air and has auniform charge distribution PSt 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 withEq. (4.25).

(c) Apply Gauss' law to calculate the total charge from Eq. (4.29)

d d·

Pv = V ·D= i1x(2x+2y) + ay(3x-2y) =0.

CHAP1ER 4160

(b) Integrate the charge density over the volume as in Eq. (4.27):

Q= {V.Dd'll= {2 {2 (20dxdydz=O.J'1-' Jx=oJy=oJz=o

With R = hi cose. we integrate from y = 0 to dJ2, which corresponds to e = 0 toeo = sin-I [(d/2}J(h2 + (dJ2f)I/2J. Thus,

ld/2 ~ Ps ld/2 cosO A Ps Leo cos20 hE= dE=z- --dy=z- --·--dO

o 1t£o 0 R 1reo 0 h cos2eA Ps n= z-t1O·

1reo

Q = f D· ds = Ffront +Fback +Fright +Fieft +F;.op + Fbonom,

Ffront= 12£2 (i2(X+Y)+Y(3X-2Y»/ . (idzdy)y=0 2=0 =2

=!.£.2(x+ytd.dy= (2+y+~Y') [JL~24,

For an infinitely wide sheet, eo = 1t/2and E = z ~ ' which is identical with Eq.(4.25).

detennine

(a) pv by applying Eg. (4.26),(b) the total charge Q enclosed in a cube 2 m on a side, located in the first octant

with three of its sides coincident with the X-, y-, and z-axes and one of itscorners at the origin, and

(c) the totaJ chroge Q in the cube, obtained by applying Eq. (4.29).

Solution:

(a) By applying Eq. (4.26)

~blem ~ Given the electric flux density

D =X2(x+ y} +y(3x- 2y) (C/m2),

CHAPTER 4

·ds =24- 8-4- 12+0+0 =O.

4.21 Repeat Problem 4.20 for D = x.xy2z3 (C/m2).

From Eq. (4.26), Pv = V. D = !(x/Z3) = lZ3.fiX •(b) Total charge Q is given by Eq. (4.27);

2 2 2 xY 4j2 2 12Q= fV.Dd'll= f f f Iz3dxdydz=-+ =64c.J'lI Jz=oJv=oJx=o L 3

- x=O .\=0 Z=O

161

From Eq. (4.15), we know a linear, isotropic material has the constitutive relationshipD = EE. Thus, we find E from D.

CHAPTER 4

Is D· ds = Ffront +Fback +Fright +Fieft +~op +Fbottom·

(c) Using Gauss' law we have

162

f 64 64Thus Q= D·ds= 3+0+0+0+0+0= 3 c.

Note that D =W;n so only Ffront and Fback (integration over z surfaces) wiII contributeto the integral.

Ffront = [2 r2 (ull)1 . (idydz)Jz=oJy=o - x=2

=LLxlzt dydz=H~nrJL = ~,

Fback = r2 r2 (Xxl~)1 . (-xdydz) = - L2 [2 xr~1 dydz = O.Jr-O ) y=o <:=0} y=ox=O x=O

Problem 4.22 Charge QI is unifonnIy distributed over a thin spherical shell ofra IUS a, and charge Q2 is unifonnly distriroted over a second spherical shell ofradius b, with b > a. Apply Gauss's law to find E in the regions R < a, a < R < b,andR> b.

Solution: Using symmetry considerations, we know D = RDR. From Table 3;1,ds = RR2sineded<p for an element of a spherical surface. Using Gauss's law inintegral form (Eq. (4.29»,

iD.dS=Qtoh

where Qtot is the total charge enclosed in S. For a spherical surface of radius R,

163

(VIm).

A RQtotE = RER = -2- =0 (VIm).4nR E

(a) In the region R < a,

Qtot = 0,

(b) In the region a < R < b,

(c) In the region R > b,

CHAPTER 4

Problem 4.23 The electric fiux density inside a dielectric sphere of radius acentered at the origin is given by

where Po is a constant. Find the total charge inside the sphere.

Solution:

Q = 1D. ds = Fe [2ft itpoR. itR2 sin 9d9d</t1JS 16=0141=0 R=a

= 21tpoa3101t sin9d6 = -21tPo~cosel~ =4npoa3 (C).

Problem 4.24 In a certain region of space, the charge density is given in cylindricalcoordinates by the function:

Apply Gauss's law to find D.

Solution: