Problem 4.48 With reference to Fig. 4-19, find E1 if E2 = x3 y2+ z2 (V/m),1 = 20, 2 = 180, and the boundary has a surface charge densitys = 3.541011 (C/m2). What angle does E2 make with the z-axis?Solution: We know that E1t = E2t for any 2 media. Hence, E1t = E2t = x3 y2.Also, (D1D2) n = s (from Table 4.3). Hence, 1(E1 n) 2(E2 n) = s, whichgives

E1z =s + 2E2z

1=

3.54101120

+18(2)

2=

3.54101128.851012 +18 = 20 (V/m).

Hence, E1 = x3 y2+ z20 (V/m). Finding the angle E2 makes with the z-axis:

E2 z = |E2|cos , 2 =

9+4+4cos , = cos1(

217

)= 61.

Problem 4.56 Figure P4.56(a) depicts a capacitor consisting of two parallel,conducting plates separated by a distance d. The space between the plates containstwo adjacent dielectrics, one with permittivity 1 and surface area A1 and anotherwith 2 and A2. The objective of this problem is to show that the capacitance C of theconfiguration shown in Fig. P4.56(a) is equivalent to two capacitances in parallel, asillustrated in Fig. P4.56(b), with

C = C1 +C2 (19)where

C1 =1A1

d (20)

C2 =2A2

d (21)

To this end, proceed as follows:(a) Find the electric fields E1 and E2 in the two dielectric layers.(b) Calculate the energy stored in each section and use the result to calculate C1

and C2.(c) Use the total energy stored in the capacitor to obtain an expression for C. Show

that (19) is indeed a valid result.

(a)

(b)

1

A1 A2

2d

+

V

C1 C2V

+

Figure P4.56: (a) Capacitor with parallel dielectricsection, and (b) equivalent circuit.

Solution:

V+

-

(c)

E1 E2

2

d

1

Figure P4.56: (c) Electric field inside of capacitor.

(a) Within each dielectric section, E will point from the plate with positive voltageto the plate with negative voltage, as shown in Fig. P4-56(c). From V = Ed,

E1 = E2 =Vd .

(b)We1 =

12

1E21 V =12

1V 2

d2 A1d =12

1V 2A1d .

But, from Eq. (4.121),We1 =

12

C1V 2.

Hence C1 = 1A1d . Similarly, C2 = 2

A2d .(c) Total energy is

We = We1 +We2 =12

V 2

d (1A1 + 2A2) =12

CV 2.

Hence,C = 1A1d +

2A2d = C1 +C2.

Problem 5.12 Two infinitely long, parallel wires are carrying 6-A currents inopposite directions. Determine the magnetic flux density at point P in Fig. P5.12.

I2 = 6 AI1 = 6 A

0.5 m

2 m

P

Figure P5.12: Arrangement for Problem 5.12.

Solution:

B = f f fm 0I1

2p (0.5) +f

f

f

m 0I22p (1.5) =

f

f

f

m 0p

(6+2) = f f f 8 m 0p

(T).

Problem 5.14 Two parallel, circular loops carrying a current of 40 A each arearranged as shown in Fig. P5.14. The first loop is situated in the xy plane with itscenter at the origin, and the second loops center is at z = 2 m. If the two loops havethe same radius a = 3 m, determine the magnetic field at:

(a) z = 0(b) z = 1 m(c) z = 2 m

z = 2 m

0

z

y

x

a

a

I

I

Figure P5.14: Parallel circular loops of Problem 5.14.

Solution: The magnetic field due to a circular loop is given by (5.34) for a loop inthe xy plane carrying a current I in the +f f f -direction. Considering that the bottomloop in Fig. is in the xy plane, but the current direction is along f f f ,

H1 =z Ia2

2(a2 + z2)3/2,

where z is the observation point along the z-axis. For the second loop, which is at aheight of 2 m, we can use the same expression but z should be replaced with (z2).Hence,

H2 =z Ia2

2[a2 +(z2)2]3/2 .

The total field is

H = H1 +H2 =z Ia2

2

[1

(a2 + z2)3/2+

1[a2 +(z2)2]3/2

]A/m.

(a) At z = 0, and with a = 3 m and I = 40 A,

H =z 4092

[133 +

1(9+4)3/2

]=z10.5 A/m.

(b) At z = 1 m (midway between the loops):

H =z 4092

[1

(9+1)3/2+

1(9+1)3/2

]=z11.38 A/m.

(c) At z = 2 m, H should be the same as at z = 0. Thus,

H =z10.5 A/m.

Problem 4.49 An infinitely long conducting cylinder of radius a has a surfacecharge density s. The cylinder is surrounded by a dielectric medium with r = 4and contains no free charges. The tangential component of the electric field in theregion r a is given by Et =cos/r2. Since a static conductor cannot have anytangential field, this must be cancelled by an externally applied electric field. Findthe surface charge density on the conductor.

Solution: Let the conducting cylinder be medium 1 and the surrounding dielectricmedium be medium 2. In medium 2,

E2 = rEr 1r2

cos ,

with Er, the normal component of E2, unknown. The surface charge density is relatedto Er. To find Er, we invoke Gausss law in medium 2:

D2 = 0,

or1r

r (rEr)+

1r

( 1

r2cos

)= 0,

which leads to r (rEr) =

(1r2

cos)

= 1r2

sin .

Integrating both sides with respect to r,

r (rEr) dr =sin 1

r2dr

rEr =1r

sin ,or

Er =1r2

sin .Hence,

E2 = r1r2

sin .According to Eq. (4.93),

n2 (D1D2) = s,

where n2 is the normal to the boundary and points away from medium 1. Hence,n2 = r. Also, D1 = 0 because the cylinder is a conductor. Consequently,

s =r D2|r=a=r 2E2|r=a=r r0

[r

1r2

sin]

r=a

=40a2

sin (C/m2).

Problem 4.50 If E = R150 (V/m) at the surface of a 5-cm conducting spherecentered at the origin, what is the total charge Q on the spheres surface?Solution: From Table 4-3, n (D1D2) = s. E2 inside the sphere is zero, since weassume it is a perfect conductor. Hence, for a sphere with surface area S = 4pia2,

D1R = s, E1R =s0

=Q

S0,

Q = ERS0 = (150)4pi(0.05)20 = 3pi02 (C).

Problem 4.58 The capacitor shown in Fig. P4.58 consists of two parallel dielectriclayers. Use energy considerations to show that the equivalent capacitance of theoverall capacitor, C, is equal to the series combination of the capacitances of theindividual layers, C1 and C2, namely

C = C1C2C1 +C2

(22)

whereC1 = 1

Ad1

, C2 = 2Ad2

(a) Let V1 and V2 be the electric potentials across the upper and lower dielectrics,respectively. What are the corresponding electric fields E1 and E2? Byapplying the appropriate boundary condition at the interface between the twodielectrics, obtain explicit expressions for E1 and E2 in terms of 1, 2, V , andthe indicated dimensions of the capacitor.

(b) Calculate the energy stored in each of the dielectric layers and then use the sumto obtain an expression for C.

(c) Show that C is given by Eq. (22).

(a)

(b)

V+

C1

C2

+

d1

d2 V

A

1

2

Figure P4.58: (a) Capacitor with parallel dielectriclayers, and (b) equivalent circuit (Problem 4.58).

Solution:

+

-

VE1

E2

1

1

V1+

-

V1+

-

d1

d2

Figure P4.58: (c) Electric fields inside of capacitor.

(a) If V1 is the voltage across the top layer and V2 across the bottom layer, then

V = V1 +V2,

andE1 =

V1d1

, E2 =V2d2

.

According to boundary conditions, the normal component of D is continuous acrossthe boundary (in the absence of surface charge). This means that at the interfacebetween the two dielectric layers,

D1n = D2n

or

1E1 = 2E2.

Hence,V = E1d1 +E2d2 = E1d1 +

1E12

d2,

which can be solved for E1:E1 =

V

d1 +12

d2.

Similarly,E2 =

V

d2 +21

d1.

(b)

We1 =12

1E21 V 1 =12

1

V

d1 +12

d2

2

Ad1 = 12 V2[

122 Ad1(2d1 + 1d2)2

],

We2 =12

2E22 V 2 =12

2

V

d2 +21

d1

2

Ad2 = 12 V2[

21 2Ad2(1d2 + 2d1)2

],

We = We1 +We2 =12

V 2[

122 Ad1 + 21 2Ad2(1d2 + 2d1)2

].

But We = 12 CV2, hence,

C = 122 Ad1 + 21 2Ad2

(2d1 + 1d2)2= 12A

(2d1 + 1d2)(2d1 + 1d2)2

=12A

2d1 + 1d2.

(c) Multiplying numerator and denominator of the expression for C by A/d1d2, wehave

C =

1Ad1

2Ad21Ad1

+2Ad2

=C1C2

C1 +C2,

whereC1 =

1Ad1

, C2 =2Ad2

.

Problem 5.9 The loop shown in Fig. P5.9 consists of radial lines and segments ofcircles whose centers are at point P. Determine the magnetic field H at P in terms ofa, b, q , and I.

b

a P

I

Figure P5.9: Configuration of Problem 5.9.

Solution: From the solution to Example 5-3, if we denote the z-axis as passing outof the page through point P, the magnetic field pointing out of the page at P due tothe current flowing in the outer arc is Houter = zI q /4p b and the field pointing outof the page at P due to the current flowing in the inner arc is Hinner = zI q /4p a. Theother wire segments do not contribute to the magnetic field at P. Therefore, the totalfield flowing directly out of the page at P is

H = Houter +Hinner = zI q4p

(1a 1b

)= z

I q (ba)4p ab .

Problem 5.10 An infinitely long, thin conducting sheet defined over the space0 xw and y is carrying a current with a uniform surface current densityJs = y5 (A/m). Obtain an expression for the magnetic field at point P = (0,0,z) inCartesian coordinates.

Solution:

z

x

x

R

P = (0, 0, z)

0

|R| = x2 + z2

w

Figure P5.10: Conducting sheet of width w in xy plane.

The sheet can be considered to be a large number of infinitely long but narrow wireseach dx wide lying next to each other, with each carrying a current Ix = Js dx. Thewire at a distance x from the origin is at a distance vector R from point P, with

R =xx+ zz.

Equation (5.30) provides an expression for the magnetic field due to an infinitely longwire carrying a current I as

H =Bm 0

=f

f

f I2p r

.

We now need to adapt this expression to the present situation by replacing I withIx = Js dx, replacing r with R = (x2 +z2)1/2, as shown in Fig. P5.10, and by assigningthe proper direction for the magnetic field. From the BiotSavart law, the directionof H is governed by lR, where l is the direction of current flow. In the present case,l is in the y direction. Hence, the direction of the field is

lR|lR| =

y (xx+ zz)|y (xx+ zz)| =

xz+ zx

(x2 + z2)1/2.

Therefore, the field dH due to the current Ix is

dH = xz+ zx(x2 + z2)1/2

Ix2p R

=(xz+ zx)Js dx2p (x2 + z2)

,

and the total field is

H(0,0,z) = w

x=0(xz+ zx)

Js dx2p (x2 + z2)

=Js2p

wx=0

(xz+ zx)dx

x2 + z2

=Js2p

(xz w

x=0

dxx2 + z2

+ z w

x=0

x dxx2 + z2

)

=Js2p

(xz

(1z

tan1(

x

z

))w

x=0+ z

( 12 ln(x

2 + z2))w

x=0

)

=5

2p

[x2p tan1

(w

z

)+ z 12(ln(w

2 + z2) ln(0+ z2))]

for z 6= 0,

=5

2p

[x2p tan1

(w

z

)+ z 12 ln

(w2 + z2

z2

)](A/m) for z 6= 0.

An alternative approach is to employ Eq. (5.24a) directly.

Problem 5.11 An infinitely long wire carrying a 25-A current in the positivex-direction is placed along the x-axis in the vicinity of a 20-turn circular loop locatedin the xy plane (Fig. P5.11). If the magnetic field at the center of the loop is zero,what is the direction and magnitude of the current flowing in the loop?

1 m

d = 2 m

I1

x

Figure P5.11: Circular loop next to a linear current(Problem 5.11).

Solution: From Eq. (5.30), the magnetic flux density at the center of the loop due to

I2

Figure P5.11: (b) Direction of I2.

the wire isB1 = z

m 02p d I1

where z is out of the page. Since the net field is zero at the center of the loop, I2 mustbe clockwise, as seen from above, in order to oppose I1. The field due to I2 is, fromEq. (5.35),

B = m 0H =z m 0NI22a .Equating the magnitudes of the two fields, we obtain the result

NI22a

=I1

2p d ,

or

I2 =2aI1

2p Nd =125

p 202 = 0.2 A.

Problem 5.13 A long, East-Westoriented power cable carrying an unknowncurrent I is at a height of 8 m above the Earths surface. If the magnetic flux densityrecorded by a magnetic-field meter placed at the surface is 15 m T when the current isflowing through the cable and 20 m T when the current is zero, what is the magnitudeof I?

Solution: The power cable is producing a magnetic flux density that opposes Earths,own magnetic field. An EastWest cable would produce a field whose direction atthe surface is along NorthSouth. The flux density due to the cable is

B = (2015) m T = 5 m T.

As a magnet, the Earths field lines are directed from the South Pole to the NorthPole inside the Earth and the opposite on the surface. Thus the lines at the surface arefrom North to South, which means that the field created by the cable is from Southto North. Hence, by the right-hand rule, the current direction is toward the East. Itsmagnitude is obtained from

5 m T = 5106 = m 0I2p d =

4p 107I2p 8 ,

which gives I = 200 A.