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**D1.1** Given points M(-1,2,1),N(3,-3,0) and P(-2,-3,-4),find (a) RMN
(b) RMN + RMP (c) |rM| (d) aMP (e) |2rP-3rN|.

SOLUTION: (a) RMN RMN = rN - rM RMN = (3x-3ay+0az)-(-1ax+2ay+1az) M(-1,2,1) N(3,-3,0) rM rN = (3+1)ax +(-3-2)ay +(0-1)az RMN = 4ax-5ay-az (b) RMN + RMP = ? RMP = rP -rM M(-1,2,1) rM = (-2,-3,-4) –(-1,2,1) P(-2,-3,-4) = (-2+1)ax +(-3-2)ay+(-4-1)az = -ax-5ay-5az We know RMN = 4ax-5ay-az RMN + RMP =(4ax-5ay-az)+ (-ax-5ay-5az) = 3ax-10ay-6az (c) |rM| =? RM = (-ax+2ay+az) |rM| = (-1)2 + (2)2+(1)2 = √6 |rM| = 2.45 (d) aMP = ? aMP = RMP |RMP| RMP = -ax-5ay-5az ______________ |RMP| = (-1)2 + (-5)2+(-5)2 ___ = √51 = 7.14

aMP = -ax-5ay-5az/7.14 =-0.14ax-0.7ay-0.7az (e) |2rP-3rN|=? Professor Aatif Saeed Preston University Islamabad rP Page 1 rP = -2ax – 3ay-4az 2rP = -4ax-6ay-az rN = 3ax -3ay+0az 3rN = 9ax-9ay+0az 2rP-3rN = (-4ax-6ay-az) – (9ax-9ay+0az) = -13ax+3ay-8az ______________ |2rP-3rN|= (-13)2 + (3)2+(8)2 =√ 242 |2rP-3rN| = 15.56 ANSWER

**D1.2.** A vector field S is expressed in rectangular coordinates as S = {125/
[(x
−1)2+(y−2)2+(z+1)2]}{(x −1)ax +(y−2)ay +(z+1)az}. (a) EvaluateS at P(2, 4,
3). (b)
Determine a unit vector that gives the direction of S at P.

**SOLUTION:
**(a) S = {125/[(x −1)2+(y−2)2+(z+1)2]}{(x −1)ax +(y−2)ay +(z+1)az}
S = {125/[(2 −1)2+(4−2)2+(3+1)2]}{(2 −1)ax +(4−2)ay +(3+1)az}
S = {125/[1+4+16]}{(ax +2ay +4az}
S = {125/21}{(ax +2ay +4az}
S =5.95{(ax +2ay +4az}
S=5.95 ax +11.9ay +23.8
(b)
aS =
S
│S│
2 5
0 2

2 5 0 2S = (5.95)2+(11.9)2+(23.8)2

=35.40+141.61+556.96 =733.97

2 5 0 2

2 5 0 2S = 27.079

= 5.95 ax +11.9ay +23.8az /27.09 = 0.219ax+0.436ay+0.878az ANSWER

**D1.3** The three vertices of a triangle are located at A(6,–1,2),B(–2,3,–4),
and C(–3,1,5). Find: (a) RAB; (b) RAC;
(c) an angle θBAC at vertex A; (d) the vector projection of RAB on RAC.

SOLUTION: (a)RAB = rB -rA R AB (2a x 3a y 4a z ) (6a x a y 2a z ) 8a x 4a y 6a z (b) RAC =rC -rA Professor Aatif Saeed Preston University Islamabad Page 2 9a x 2a y 3a z (c) R AB R AC R AB R AC cosBAC cos BAC R AB R AC R AB R AC (8a x 4a y 6a z ) (9a x 2a y 3a z ) (8)2 (4) 2 (6) 2 (9) 2 (2) 2 (3) 2 0.594 62 116 94 BAC cos 1 (0.594) 53.56 (d) R AB on R AC R AB a AC a AC ( 9a x 2a y 3a z ) ( 8 a 4 a 6 a ) x y z ( 9) 2 (2) 2 (3) 2 ( 9a x 2a y 3a z ) 2 2 2 ( 9) (2) (3) 62 ( 9a x 2a y 3a z ) 94 94 5.963a x 1.319a y 1.979a z ANSWER

**D1.4.** The three vertices of a triangle are located at A(6,−1, 2), B(−2, 3,
−4),and C(−3, 1, 5). Find:
(a) RAB × RAC; (b) the area of the triangle; (c) a unit vector perpendicular
to the plane in
which the triangle is located.

SOLUTION: A (a) RAB = rB -rA RAB = (-2,3,-4)-(6,-1,2) =(-2-6),(3+1),(-4-2) = -8,4,-6 RAB rA B C rB RAB = -8ax+4ay-6az Professor Aatif Saeed Preston University Islamabad Page 3 RAC =rC -rA RAC = (-3,1,5)-(6,-1,2) = (-3-6),(1+1),(5-2) = -9,2,3 A RAc = -9ax+2ay+3az RAB RAC = (b) C ax ay az -8 4 -6 -9 2 3 = ax(4x3 -2(-6))-ay(-8x3 –(-6)(-9))+az(2(-3)-4(-9)) =ax(12+12)-ay(-24-54)+az(-16+36) = 24ax -78ay+20az Area =1/2│ RAB × RAC│

2 5 0 2

2 5 0 2= 1/2 24ax -78ay+20az rA

rC B A =1/2(24)2 + (-78)2 + (20)2 B C =1/2576+6084+400 =1/2 7060 =1/2(84.0) = 42.0 (c) Unit vector perpendicular to the plane in which triangle is located =? Unit vector of RAB × RAC = RAB × RAC │ RAB × RAC│ = 24ax -78ay+20az /84.0 = 0.286ax+0.928ay+0.238az ANSWER

** D1.5.** (a) Give the rectangular coordinates of the point C( 0 3C 1 0 3C 6 = 4.4, =
2 5
E 6−115 , z = 2).

(b) Give the cylindrical coordinates of the point D(x =−3.1, y = 2.6, z = −3).

(c) Specify the distance from C to D.

Solution:

(a) polar coordinates into rectangular coordinates Given: 0 3 C 1

0 3 C 6

2 5 E 6 = 4.4, =−115 , z = 2

And x = ρcos φ y = ρsin φ z = z so x = 4.4cos(-1150) x = -1.860 y = 4.4sin(-1150) y = -3.99 z=2 therefore rectangular coordinates of point C are x = -1.860, y = -3.99, z = 2 C (x = -1.860, y = -3.99, z = 2) (b) Rectangular coordinates into cylindrical coordinates Given: x = − 3.1, y = 2.6, z = − 3 Professor Aatif Saeed Preston University Islamabad Page 4 And 0 3C 1 = x2+y2 0 3 C 6 = tan-1y/x z = z Now 0 3 C 1 = (-3.1)2+(2.6)2 = 9.61+6.76 = 16.37 0 3 C 1 = 4.05 0 3 C 6 = tan-1(2.6/-3.1) 0 3 C 6 = -39.90 Angle is in II quadrant so

0 3 C 6 = 1800-39.90 0 3 C 6 = 1400 z = -3 therefore cylindrical coordinates of point D are 0 3C 1 0 3C 6 = 4.05, = -3.990, z = -3 D( 0 3C 1 0 3C 6 = 4.05, = -3.990, z = -3) (c) Distance from C to D C = -1.860ax-3.99ay+2az point D(x =−3.1, y = 2.6, z = −3) RCD = rD - rC = ( -3.1ax-2.6ay-3az)- (-1.860ax-3.99ay+2az) = -1.24ax+6.59ay-5az 2 5 0 2

2 5 0 2

2 5 0 2

2 5 0 2RCD = (-1.24)2+(6.59)2+(-5)2 RCD =8.36 ANSWER

**D1.6.** Transform to cylindrical coordinates: (a) F = 10ax −8ay+6az at point
P(10,−8, 6); (b)G =
(2x+y)ax −(y−4x)ay at point Q( 0 3C 1 0 3C 6, , z). (c) Give the rectangular
components of the vector H =
20a 0 3C 1 − 10a 0 3C 6 + 3az at P(x = 5,y = 2, z = −1).

**Solution:** (a) F = 10ax−8ay+6az
FP = F.aP = (10ax−8ay+6az).aP
= 10ax.aP-8ay.aP+6az.aP
= 10cos 0 3C 6-8sin 0 3C 6+0 (1) For point P, x = 10 ,y = -8
0 3
A 6 = tan-1y/x
= tan-1(-8/10)
= -38.65980
As y is –ve and x is +ve 0 3C 6 is in the 4th quadrant.hence 0 3C 6 calculated is
correct
Cos 0 3C 6=0.7808 ,sin 0 3C 6= -0.6246 (1) FP = 10(0.7808)-8(-0.6246)
= 12.804aP
F 0 3C 6 = F.a 0 3C 6 = (10ax−8ay+6az).aφ
=10ax.a 0 3C 6−8ay.a 0 3C 6+6az.aφ
= 10(-sinφ)-8(cosφ)+0
= 10(-0.6246)-(0.7808)
= -6.246+6.246 Professor Aatif Saeed Preston University Islamabad Page
5 F 0 3C 6 = 0
FZ= F.az = (10ax−8ay+6az).az

=10ax.az−8ay.az+6az.az = 0+0+6az Hence F = 12.4aP+6az (b) G = (2x+y)ax −(y−4x)ay GP =G.aP =( (2x+y)ax −(y−4x)ay ).aP = (2x+y)ax.aP−(y−4x)ay.aP = (2x+y)cos 0 3C 6−(y−4x)sinφ =(2ρcosφ+ ρsin 0 3C 6)cosφ−(ρsinφ−4ρcos 0 3C 6)sinφ = 2ρcos2φ+ ρsinφcosφ−ρsin2φ+4ρcosφsinφ =( 2ρcos2φ+ 5ρsinφcosφ−ρsin2 0 3C 6)aP G 0 3C 6 =G.a 0 3C 6 =( (2x+y)ax −(y−4x)ay ).aφ = (2x+y)ax.a 0 3C 6−(y−4x)ay.aφ = -(2x+y) sin 0 3C 6 −(y−4x) cosφ =-(2ρcosφ+ ρsin 0 3C 6) sinφ −(ρsinφ−4ρcos 0 3C 6)cosφ = (4ρcos2φ−ρsin2φ-3ρcosφsin 0 3C 6)aφ Gz =G.az =( (2x+y)ax −(y−4x)ay ).az = (2x+y)ax.az−(y−4x)ay.az =0 Hence G =( 2ρcos2φ+ 5ρsinφcosφ−ρsin2φ)aP ++(4ρcos2φ −ρsin2φ-3ρcosφsinφ)aφ H = 20a 0 3C 1 − 10a 0 3C 6 + 3az Hx = H.ax = (20a 0 3C 1 − 10a 0 3C 6 + 3az).ax = 20a 0 3C 1.ax − 10a 0 3C 6.ax + 3az.ax = 20cos 0 3C 6+10sinφ At point P , x = 5 , y = 2 ,z = -1 0 3 A 6 = tan-1y/x = tan-1(2/5) = 21.80140 Cos 0 3C 6=0.9284 ,sin 0 3C 6= -0.3714 Hx = 20(0.9284) + 10(0.3714) = 22.282 Hy = H.ay = (20a 0 3C 1 − 10a 0 3C 6 + 3az).ay = 20a 0 3C 1.ay − 10a 0 3C 6.ay + 3az.ay = 20sin 0 3C 6-10cosφ

= 20(0.3714) -10(0.9284) = -1.856 Hz = H.az = (20a 0 3C 1 − 10a 0 3C 6 + 3az).az = 20a 0 3C 1.az − 10a 0 3C 6.az + 3az.az (c) Professor Aatif Saeed Preston University Islamabad Page 6 = 0+0+3 =3 Hence H = (22.3,-1.857,3) ANSWER

**D1.7**. Given the two points, C(−3, 2, 1) and D(r =5, θ =20◦, φ= −70◦),find:
(a) the
spherical coordinates of C; (b) the rectangular coordinates of D;
(c) the distance from C to D.

**Solution:
**(a) C = -3,2,1 r = x2+y2+z2
=-32+22+12
= 3.742 =
cos-1
z
x2+y2+z2
= cos-1 (1/3.742) = 74.50
0 3
C 6 = tan-1y/x
0 3
C 6 = tan-12/-3 = -33.690+1800
= 146.310
Hence C(r=3.742, =74.500 , 0 3C 6 =146.310)
(b) D(r =5, θ =20◦, φ= −70◦)
X = rsincosφ
0 = 5sin(20)cos(-70 )
=5(0.342)(0.342)
= 0.585 Z = rcos
= 5cos(200)
= 5(0.939)
= 4.698 y = rsinsinφ
=5sin(20)sin(-700)
=5(0.342)(-0.939)

= -1.60 Hence D (x = 0.585,y = -1.607,z = 4.698) (c) Distance from C to D C = -3ax+2ay+az D =0.585ax-1.607ay+4.70az RCD = rD - rC = (0.585ax-1.607ay+4.70az)- (-3ax+2ay+az) = 3.585ax-3.607ay+3.7az 2 5 0 2

2 5 0 2

2 5 0 2

2 5 0 2RCD = (3.585)2+(-3.607)2+(3.7)2 RCD =6.289 6.29 ANSWER

**D2.1**. A charge QA = −20µC is located at A(−6, 4, 7), and a charge QB =
50µC is at
B(5, 8,−2) in free space. If distances are given in meters, find:

2 5 0 2

2 5 0 2(a) RAB; (b) RAB . Determine the vector force exerted on QA by QB if €

0 =: (c) 109/(36π) F/m; (d) 8.854 × 10-12 F/m.

**Solution:
**(a) RAB = rB - rA
Professor Aatif Saeed Preston University Islamabad Page 7 = (5-(-6))ax
+(8-4)ay +(-2-7)az

2 5 0 2

2 5 0 2= 11ax+4ay -9az (b) RAB =(11)2+(4)2+(-9)2 = 14.76m

(c) FAB = QAQBRAB 2 5 0 2

2 5 0 24o RAB 3

2 5 0 2

2 5 0 2= ( -20 10-6)(5010-6)( 11ax+4ay -9az )/4(10-9/36) 14.7 3

2 5 0 2

2 5 0 2= -9 103 10-610-6109( 11ax+4ay -9az ) 14.7 3

=30.76ax+11.184ay -25.16az mN (d) FAB = QAQBRAB

2 5 0 2

2 5 0 2

2 5 0 2

2 5 0 24o RAB 3 = ( -20 10-6)(5010-6)( 11ax+4ay -9az )/48.8510-12 14.7 3

2 5 0 2

2 5 0 2= -103 10-121012( 11ax+4ay -9az) /48.85 14.7 3

=30.72ax+11.169ay -25.13az Mn ANSWER

**D2.2.** A charge of −0.3µC is located at A(25,−30, 15) (in cm), and a second
charge of
0.5µC is at B(−10, 8, 12) cm. Find E at: (a) the origin; (b) P(15, 20, 50) cm.

**Solution:
**(a) Let E at the origin is denoted by E0 and it will be the sum of EA(E due

to QA located at point A) and EB(E due to QB located at point B) EA = QAROA

2 5 0 2

2 5 0 24o ROA 3 ROA = (0-(25))ax +(0-(-30))ay +(0-15)az cm

= -25ax+30ay -15az 2 5 0 2

2 5 0 2ROA =(-25)2+(30)2+(-15)2

=41.83cm 2 5 0 2

2 5 0 2EA = (-0.310-6)( -25ax+30ay -15az)/4 8.85 10-12 41.83 10-2 3

= -368.55(-25ax+30ay -15az) EB = QBROB

2 5 0 2

2 5 0 24o ROB 3

ROB = (0-(-10))ax +(0-8)ay +(0-12)az cm = 10ax-8ay -12az 2 5 0 2

2 5 0 2ROB =(10)2+(-8)2+(-12)2

=17.55cm 2 5 0 2

2 5 0 2EB = (0.510-6)( 10ax-8ay -12az)/4 8.85 10-12 17.55 10-2 3

= 8317.36(10ax-8ay -12az ) E = EA + EB = -368.55(-25ax+30ay -15az)+ 8317.36(10ax-8ay -12az) =92,3ax -77.6ay -94.2az KV/m (b) Now at point P RPA = (15-25)ax +(20-(-30))ay +(50-15)az = -10ax+50ay+35az Professor Aatif Saeed Preston University Islamabad

2 5 0 2

2 5 0 2Page 8 RPA = (-10)2+(50)2+(35)2

=61.84 EPA = QARPA 2 5 0 2

2 5 0 24o RPA 3

2 5 0 2

2 5 0 2= (-0.310-6)( -10ax+50ay +35az)/4 8.85 10-12 61.84 10-2 3

= -114.003( -10ax+50ay +35az) RPB = (15-(-10))ax +(20-8)ay +(50-12)az = 25ax+12ay+38az 2 5 0 2

2 5 0 2RPB =(25)2+(12)2+(38)2

=47.04 EPB = QBRPB

2 5 0 2

2 5 0 24o RPB 3

2 5 0 2

2 5 0 2= (0.510-6)( 25ax+12ay +38az)/4 8.85 10-12 47.04 10-2 3

= 431.54(25ax+12ay +38az) Er = EPA +EPB =-114.003( -10ax+50ay +35az)+ 431.54(25ax+12ay +38az) = 11.9ax -0.519ay +12.4az KV/m ANSWER

**D2.3.**Evaluate the sum: (a) ∑= + (−)m/m2+1 (b) ∑=(. ) + 2/(4+m2)1.5
SOLUTION: (a) = 0 ∶ 1 + (−1)0/02+1 = 2 = 1 ∶ 1 + (−1)1/12+1 = 0 = 2 ∶ 1 +
(−1)2/22+1 = 2/5 = 3 ∶ 1 + (−1)3/32+1 = 0 = 4 ∶ 1 + (−1)4/42+1 = 2/17 = 5 ∶
1 + (−1)5/52+1 = 0
∑5=0 1 + (−1)m/m2+1 = 2+0+2/5+0+2/17+0 = 2.52 (b)
m = 1 : (0.1)1+12/(4+12)1.5 =1.1/11.18
m = 2 : (0.1)2+12/(4+22)1.5 =1.01/22.62
m = 3 : (0.1)3+12/(4+32)1.5 =1.001/46.87
m = 4 : (0.1)4+12/(4+42)1.5 =1.0001/89.44
4 ∑=1(0.1) + 12/(4+m2)1.5 = 0.1755

**D2.4.** Calculate the total charge within each of the indicated volumes: (a)
0.1 ≤ |x|, |y|, |z| ≤ 0.2: 0 3C 10 3B D =
1/ x3 y3z3 ; (b) 0 ≤ 0 3C 1 0 3C 6 0 3C 0 ≤ 0.1, 0 ≤ ≤ , 2 ≤ z ≤ 4; 0 3C 10 3B D 0 3C 1 = 2z2 sin 0.6 0 3C 6; (c)
universe: 0 3C 10 3B D = e−2r /r 2.

**Solution:
**Professor Aatif Saeed Preston University Islamabad Page 9 0.1 0.1 0.1 0.2
0.2 0.2
3 3 Q = vol ρdv = 1/ x y z dxdydz + 1/ x y z
3 (a) 0.2 = 0.2 (b) 1 0.1 8x2│ y2│

2 5 0 20.2 0.1 0.1 _ 0.1 0.2 z2 0.2 3 0.2 1

2 5 0 2

2 5 0 2

2 5 0 20.1 3 3 = 0.2 0.2 8x2 y2 z2 0.2 0.1 0.1 4 0 3C 13z2sin0.6 φdzdρd 0 3C 6 dxdydz

0.1 1 _ 8 (0.03) 1 = 0 8 (0.03) 0.1 0.1 0 0 0 0.1 4 3 2

0 3 C 1= d 0 3C 1 z dz sin0.6dφ

0

0 0 (c) 0.1 4 = (-cos0.6 0 3C 6/0.6 2 50 2) ( 0 3C 14/4 2 50 2 2 50 2) (z3/3) =1.018 Mc 0 0 0 2 2 -2r e sindφddr 0 0 0 2 2 e 2rdr sin d d 0 3C 6 = - 0 0 0 2 2 (-e-2r/2 2 50 2) (cos )│ 0 3C 6 2 50 2( ) = 0 0 = -6.28 C 0 ANSWER

**D2.5.** Infinite uniform line charges of 5 nC/m lie along the (positive
andnegative) x and y axes in free space.
Find E at: (a) PA(0, 0, 4); (b) PB(0, 3, 4).

**Solution:
**(a) E(PA) = PLx .apx + PLy .apy 20px 20py = 5 10-9 0ax +0ay+4 az + 5
10-9
20(4) 0 +0+42 0ax +0ay+4 az 20(4) 0 +0+42 = 22.47az + 22.47az
= 44.939 az v/m
(b) E(PB) = PLx .apx
20px + PLy .apy
20py = 5 10-9 0ax +3ay+4 az + 5 10-9
Professor Aatif Saeed Preston University Islamabad 0ax +3ay+4 az
Page 10 20(5) 0 +32+42
= 5 10-9
20(5) 20(4) (0.6ay+0.8az) 0 +32+42 5 10-9 .az
20(4) + = 10.785ay +36.850az v/m ANSWER

**D2.6.** Three infinite uniform sheets of charge are located in free space
asfollows: 3 nC/m2 at z = −4, 6
nC/m2 at z = 1, and −8 nC/m2 at z = 4.Find E at the point: (a) PA(2, 5,−5);
(b) PB(4, 2,−3); (c) PC(−1,−5,
2); (d)PD(−2, 4, 5). (i) Electric field due to 3 nC/m2 : E1 = s aN = 20
s aN = (ii) Electric field due to 6nC/m2 : E1 = 20
(iii) Electric field due to -8 nC/m2: E1 = s aN = 20 3 10-9 aZ 20
610-9 20

-8 10-9 20 aZ = 169.5 aZ = 338.8 az aZ = -451.76 az According to direction of point relative to normal (a) E = - E1 – E2 – E3 = -56.6 az v/m (B) E = + E1 – E2 – E3 = 283 az v/m (c) E = + E1 + E2 – E3 = 961 az v/m (d) E = + E1 + E2 + E3 = 56.6 az v/m

**D2.7.** Find the equation of that streamline that passes through the point P
(1, 4, −2) in the field E = (a)
−8x/yax +4x2/y2 ay; (b) 2e5x [y(5x +1)ax +xay ].

**Solution :** (a) E = ( -8x/y )ax + ( 4x2/y2 )ay P (1,4,-2 ) dy/dx = Ey/Ex
dy/dx = ( -8x/y ) ( 4x2/y2 )
dy/dx = -x/2y
2ydy = -xdx
2ydy = -xdx
x2 +2y2 = c
Professor Aatif Saeed Preston University Islamabad Page 11 put x=1 and
y = -4 for value of c
(1)2 + 2 (-4)2 = c
C = 33
So x2 +2y2 = 33 (b) dy/dx = Ey /Ex dy/dx = ( y(5x + 1 ) ( x ) D3.2. Calculate
D in rectangular coordinates at point P(2,−3, 6) producedby: (a) a point
charge QA =
55 mC at Q(−2, 3,−6); (b) a uniform linecharge ρLB = 20 mC/m on the x
axis; (c) a uniform surface
charge density ρSC = 120 C/m2 on the plane z = −5 m. Solution :
Q(−2, 3,−6) and P(2,−3, 6 ) RPQ = (2 - (-2) ) ax + (-3 - 3) ay + (6 - (-6) ) az
= 4ax - 6ay + 12az
2 5
0 2

2 5 0 2RPQ = (4)2 + (-6)2 + (12)2

= 196 = 14 -3

D = (55 10 (4ax - 6ay + 12az)) (4(14)3) = 6:38ax – 9.57ay + 19.14az Cm2 L= 20mC along X axis ( X ) ( c ) E = PLx .Rpx

2 5 0 2

2 5 0 220 Rpx 2

Rpx = P(2,-3,6) - (x, 0, 0) = (2 - x)ax - 3ay + 6az 2 5 0 2

2 5 0 2RPx =

= (-3)2 + (6)2 45 D = 0E D = PLx .Rpx

2 5 0 2

2 5 0 22 Rpx 2 = 20 10-3(-3ay + 6az) 245

D = -212ay + 424az Cm2 E = = = s az 20 120 az 20 60 az Professor Aatif Saeed Preston University Islamabad Page 12 0 D = 0E = 60 az ANSWER D3.3. Given the electric flux density, D = 0.3r 2ar nC/m2 in free space:(a) find E at point P(r = 2, θ = 25◦, φ = 90◦); (b) find the total chargewithin the sphere r = 3; (c) find the total electric flux leaving the sphere r = 4. Solution: (a) E =? at point P(r = 2, = 250 , = 900) D = 0E E = D 0 (b) = (0.3 r2arnCm2) 8.85 10-12Fm = (0.3 22arnCm2) 8.85 10-12Fm = 135.5 arVm Q = ? for a sphere of radius r=3 Q = ∮ . ds = r2sindd 2

2 2 2 B

2 2 2 BQ = 0.3r2 10-9 0 0 2 = 0.3 r2(4r2) 10-9

= 1.2r410-9 = (1.2r4)r=4 10-9 = 305 nc ( c ) = Q =? for a sphere of radius r=4 = 1.2r410-9 = (1.2r4)r=4 10-9 = 965 c ANSWER

**D3.4.** Calculate the total electric flux leaving the cubical surface formed by
the six planes x, y, z = 5 if
the charge distribution is: (a) twopoint charges, 0.1 C at (1,−2, 3) and 1/7C
at (−1, 2,−2); (b) a uniform

line charge of 0 3C 0 C/m at x = −2, y = 3; (c) a uniform surface charge of 0.1 C/m2 on the plane y = 3x.

**Solution :** (a) since both the given charges are enclosed by the cubical
volume according to the gauss's law = Q1 + Q2 = 0.1C + (17)C = 0.243C
(b) L = C at (-2,3,z)
x=-2 and y=3 and is parallel to z axis, the total length of this charge
distribution enclosed by the given
cubical volume is 10 units as z = 5 so =Q = L 10 = 10 = 31.4C Professor
Aatif Saeed Preston University Islamabad Page 13 ( c ) s = 0.1 C on the
plane y=3x, now this is a straight line equation in xy plane which passes
through the origin. 103
10
y = 3x 10
The length is moving up and down along z axis between z = 5
By putting y=5 and x=5/3
L1 = (5)2 + ( 5/3)2
= 25 + 259
L1 =
5.270
The same length we will get on the plane formed by –ve x and –ve y axis
L2 = (-5)2 + (- 5/3)2
= 25 + 259
L2 =
5.270
L = 5.270 + 5.270
= 10.540
Now this straight line is moving between z = 5 to form a plane whose area
is given by
1010.540 = 105.40
surface charge density s = 0.1C
now according to gauss's law = Qenclosed = s (area of the plane) = 0.1C
(105.40) = 10.54C ANSWER

**D3.5.** A point charge of 0.25 C is located at r = 0, and uniform surface
charge densities are located as
follows: 2 mC/m2 at r = 1 cm, and −0.6 mC/m2 at r = 1.8 cm. Calculate D
at: (a) r = 0.5 cm; (b) r = 1.5
cm; (c) r = 2.5 cm. (d) What uniform surface charge density should be
established at r = 3 cm to cause D
= 0 at r = 3.5 cm?

**Solution:** (a) r = 0.5cm and Q1 = 0.25c
D = (Q4r2)ar
D = (0.25 10-64(0.5 10-2)2ar
= (b) (c) ( as r=0.5cm and Q1 = 0.25c ) 796 arC Q2 = s (area of the sphere)
D = 2 10-3 cm 4(1 10-2)2
= 2.513 10-6 c
= (Q1 + Q2 4r2)ar D = (0.25 10-6 + 2.513 10-6 4(1.5 10-2)2ar ( r = 1cm) ( r
= 1.5 cm) = 977 arC
Q3 = s (area of the sphere) = -0.6 10-3 cm 4(1.8 10-2)2
Professor Aatif Saeed Preston University Islamabad Page 14 = - 2.5 10-6 c
( r = 2.5 cm) 40.74 arC = (d) 4(2.5 10-2)2ar = (0.25 10-6 + 2.513 10-6 - 2.5
10-6 D Q4 = 4(3 10-2)2 s ( r = 3 cm) Q1 + Q2 + Q3 + 4(3 10-2)2 s
4(0.035)2 D = = -28.29 arC = 0 ANSWER

**D3.6.** In free space, let D = 8xyz4ax+4x2z4ay+16x2yz3az pC/m2. (a) Find
the total electric flux passing
through the rectangular surface z = 2, 0 < x < 2, 1 < y < 3, in the az
direction. (b) Find E at P(2,−1, 3). (c)
Find an approximate value for the total charge contained in an incremental
sphere located at P(2,−1, 3)
and having a volume of 10−12 m3.

**Solution :
**a) = ∮ . (8xyz4ax + 4x2z4ay + 16x2yz3az)dxdydz 3 2 = (8xyz4ax + 4x2z4ay
+ 16x2yz3az)dxdydz
1 0 = 3 2 16x2yz3dxdy
1 0

3 2 2 = 16z3 ydy x dx 1 = 16z 3 0 4 83z = 2 = 1365 pc (b) D = 0E E = D 0 E = (8xyz4ax + 4x2z4ay + 16x2yz3az) 10-2 8.85 10-2 X =2,y = -1, z = 3 (c) E = -1296 ax + 1296 ay + 1728 az 8.85 10-2 E = 14064 ax + 146.4 ay – 195.25 az Q=? V = 10-12m3 Dx = (8xyz4) x x = (8yz4) x = 2 ,y = -1 , z = 3 = 8(-1)(81) Professor Aatif Saeed Preston University Islamabad Page 15 = -648 Dy = (4x2z2) y y = 0 Dz = (16x2yz3) z z = -1728 Dx x + Dz z Q = + Dy y V = ( -648 – 1728 ) 10-12 10-12m3 ( p = 10-12 ) = - 2.38 -21 c ANSWER

**D3.8** .Determine an expression for the volume charge density associated
witheach D field: (a) D
=4xy/z ax + 2x2/z ay −2x2 y/z2 az

**Solution:
**D = 4xy/z ax + 2x2/z ay −2x2 y/z2 az
divD = (4xyz-1) + (2x2z-1) - (2x2yz-2)
x
y
z

-1 2 -3 = 4yz + 0 -2(-2)(x yz ) = 4y + 4 x2y z z3 = 4yz2 + 4x2y z3 2 = 4y (x + z2) z3

**D4.1.** Given the electric field E =1/z2 (8xyzax +4x2zay −4x2 yaz ) V/m, find
the differential amount of
work done in moving a 6-nC charge a distance of 2 m, starting at P(2,−2, 3)
and proceeding in the
direction aL = (a) −6/7ax + 3/7ay + 2/7az; (b) 67ax − 37
ay − 27az; (c) 37ax + 67ay .

**Solution:
**E = 1 z2 (8xyzax +4x2zay −4x2 yaz ) v/m dL = 2m (a) P(2,−2, 3) Q = 6-nc
aL = −6/7ax + 3/7ay + 2/7az
dL = aL . dL = 2m (−6/7ax +
= Q 3/7ay + 2/7az ) 2 10-6m (−6/7ax + 3/7ay + 2/7az ) = -QE.dL Professor
Aatif Saeed Preston University Islamabad Page 16 Q = (-6 10-9) 1
z28xyzax + 4x2zay – 4x2 yaz .dL = -6 10-9 210-61 z28xyzax + 4x2zay –
4x2 yaz .−6/7ax + = -1210-151 z2(8xyz)(-6/7) + (4x2z )(3/7)– (4x2 y)(2/7)
= -1210-151 z2-48/7xyz 3/7ay + 2/7az + 12/7x2z – 8/7x2 y Putting values
x = 2 , y = -2 and z = 3
= -1210-151 (3)2(-48/7)(2)(-2)(3)+ (12/7)(2)2(3) – (8/7)(2)2(-2) =
-1210-1519[ 576/7 + 144/7 + 64/7 ]
= -410-1513 [576+144+64]
7
-15

= -4/3 10 [ 112 ] = -448/3 10-15 = -149.3 10-15 = -149.3 fJ (b) E = 1 z2 (8xyzax +4x2zay −4x2 yaz ) v/m dL = 2m P(2,−2, 3) Q = 6-nc aL = 6/7ax - 3/7ay - 2/7az dL = aL . dL = 2m (6/7ax = Q 3/7ay - 2/7az ) 2 10-6m (6/7ax - 3/7ay - 2/7az ) = -QE.dL Q = (-6 10-9) 1 z28xyzax + 4x2zay – 4x2 yaz .dL = -6 10-9 210-61 z28xyzax = -1210-151 z2(8xyz)(6/7) = -1210-151 z248/7xyz + 4x2zay – 4x2 yaz .6/7ax - 3/7ay - 2/7az + (4x2z ) (-3/7)– (4x2 y)(-2/7) - 12/7x2z + 8/7x2 y Putting values x = 2 , y = -2 and z = 3 = -1210-151 (3)2(48/7)(2)(-2)(3)+ (-12/7)(2)2(3) + (8/7)(2)2(-2) = -1210-1519[ -576/7 - 144/7 - 64/7 ] Professor Aatif Saeed Preston University Islamabad Page 17 = -410-1513 [-576-144-64] 7 -15 = -4/3 10 [ -112 ] = 448/3 10-15 = 149.3 10-15 = 149.3 fJ (c) E = 1 z2 (8xyzax +4x2zay −4x2 yaz ) v/m dL = 2m P(2,−2, 3) Q = 6-nc aL = 3/7ax + 6/7ay dL aL . dL = = 2m (3/7ax + = Q 6/7ay ) 2 10-6m (3/7ax + 6/7ay + ) = -QE.dL Q = (-6 10-9) 1 z28xyzax + 4x2zay – 4x2 yaz .dL = -6 10-9 210-61 z28xyzax = -1210-151 z2(8xyz)(3/7) = -1210-151 z224/7xyz + 4x2zay – 4x2 yaz .3/7ax + 6/7ay + (4x2z )(6/7) + 24/7x2z Putting values x = 2 , y = -2 and z = 3 = -1210-151 (3)2(24/7)(2)(-2)(3)+ (24/7)(2)2(3) = -1210-1519[ -288/7 + 288/7 ] = -410-1513 [-288+288 ] 7

= -4/3 10-15 [ 0 ] = 0 J ANSWER

**D4.2.** Calculate the work done in moving a 4-C charge from B(1, 0, 0) to A
(0, 2, 0) along the path y = 2 − 2x, z = 0 in the field E = (a) 5axV/m; (b)
5xaxV/m; (c) 5xax + 5yayV/m.

**Solution :
**( a)
Q = 4C A(0, 2, 0) B(1, 0, 0) Path y = 2- 2x z=0 E = 5 ax v/m
A W = -Q E.dL
B
A Professor Aatif Saeed Preston University Islamabad Page 18 W = -4 5.dx
B Where dL = dxax+dyay+dzaz
A
= -20X
B
= -20 (0-1)
= 20 j
(b)
Q = 4C
A(0, 2, 0)
E = 5x ax v/m B(1, 0, 0) Path y = 2- 2x z=0 A W = -Q E.dL
B A W = -45x.dx
B Where dL = dxax+dyay+dzaz
A
2
= -20X /2
B
= -10 (0-1)
= 10 j (c) E = 5xax + 5yay
A W = -Q E.dL
B
A W = -4 (5xax + 5yay).dx
B Where dL = dxax+dyay+dzaz

A = -20 x2/2 + y2/2 B = -10 ( (0-1)+(4-0) ) = -10(-1+4) = -30 joules ANSWER

**D4.3.** We will see later that a time-varying E field need not be conservative.
(If it is not conservative,
the work expressed by Eq. (3) may be a function of the path used.) Let E =
yaxV/m at a certain instant of
time, and calculate the work required to move a 3-C charge from (1, 3, 5) to
(2, 0, 3) along the straightline segments joining: (a) (1, 3, 5) to (2, 3, 5) to
(2, 0, 5) to (2, 0, 3); (b) (1, 3, 5) to (1, 3, 3) to (1, 0, 3) to
(2, 0, 3).

**Solution:** (a) E =yax Q = 3-C
A W = -QE.dL
Professor Aatif Saeed Preston University Islamabad Page 19 B (2,3,5)
(2,0,5) (2,0,3)
W = -3 E.dL + E.dL + (1,3,5) (2,3,5)
Where E.dL
(2,0,5) E.dL = ( yax ) .( dxax+dyay+dzaz) = ydx ydx + ydx + ydx W = -3 A1
A2 A3 As X is only varing in the case of A1 so dx = 0 for A2 , A3
W = -3 ydx as y is a constant for A1 so y = 3 A1 = -3 (3)dx
A1 = -9 x
A1
= -9(2-1)
= -9 Joules (b) (1,3,3) (1,0,3)
W = -3 Where (2,0,3) E.dL + E.dL + E.dL
(1,3,5) (1,3,3) (1,0,3) E.dL = ydx As dx 0 for the third integral only; so the
first two integrals are zero.
(2,0,3)
So w = -3 ydx ( y = constant = 0 ) (1,0,3)

= -9(0)dx = 0 Joules ANSWER

**D4.4.** An electric field is expressed in rectangular coordinates byE =6x2ax
+ 6yay
+4azV/m. Find: (a) VMN if points M and N are specified by M(2, 6,−1) and
N(−3,−3, 2);
(b) VM if V = 0 at Q(4,−2,−35); (c) VN if V = 2 at P(1, 2,−4).

**Solution :** (a) E = 6x2ax+ 6yay +4az V/m
M = (2, 6,−1)
,
N = (−3,−3, 2)
A V = - E.dL
B
Professor Aatif Saeed Preston University Islamabad Page 20 (2,6,-1) = -
6x2dx+ 6ydy +4dz
(-3,3,2)
(2,6,-1) = - (2x3 +3y2 + 4z )
(-3,3,2) = - (2(23 – (-3)3) +3(62 –(-3)2) + 4(-1-2)
= - ( 70 +81 -12 )
= -139 V (b) VMQ = VM - VQ
VM = VMQ + VQ
So finding VMQ
M VMQ = - E.dL
Q
M = - 6x2dx+ 6ydy +4dz
Q (2,6,-1) = - (2x3 +3y2 + 4z )
(4,-2,-35) (2(23 (4)3) = –
+3(62 –(-2)2) + 4(-1+35)
= - ( -112 +96 +136 )
= -120 V
But given that VQ = 0 VM = VMQ = -120V
(c) VNP = VN - VP
VN = VNP + VP

VN = VNP + 2 So finding VNP ( VP = 2 ) N VNP = - E.dL P M = - 6x2dx+ 6ydy +4dz Q (-3,3,2) = - (2x3 +3y2 + 4z ) (1,2,-4) (2(-3)3 = –1) +3(32 - 22) + 4(2+4) = - ( -56 +15 +24 ) Professor Aatif Saeed Preston University Islamabad Page 21 = 17V But given that VP = 2 VN = 17 + 2= 19 Volts ANSWER

**D4.5.** A 15-nC point charge is at the origin in free space. Calculate V1 if
pointP1 is located at
P1(−2, 3,−1) and (a) V = 0 at Q (6, 5, 4); (b) V = 0 at infinity;(c) V = 5 V at
(2, 0, 4).

**Solution:** (a) Q = 15-nc at origin P1 = (−2, 3,−1) V = 0 at Q (6, 5, 4)
As we know
VAB = Q/40 [ 1/rA - 1/rB ]
Is the voltage of point A with reference to point B
rA = 22 + 32 + 1 = 14 rB = 62 + 52 + 42 = 77 VAB = 15 10-9/40 [ 1/ 14 - 1/
14 ]
VAB = 20.67 V
VAB = VA - VB
VPQ = VP - VQ
So
(b) (VB = VQ = 0) VPQ = VP = 20.67 V
Q = 15-nc at origin P1 = (−2, 3,−1) V = 0 at infinty
As we know
VPQ = Q/40 [ 1/rA - 1/rB ]
rA = 22 + 32 + 1 = 14 rB = Professor Aatif Saeed Preston University
Islamabad Page 22 VPQ = 15 10-9/40 [ 1/ 14 - 1/ ]
VPQ = 36 V
(b) Q = 15-nc at origin P1 = (−2, 3,−1) VQ = 5V at Q(2, 0, 4)
As we know
VPQ = Q/40 [ 1/rA - 1/rB ]

rA = 14 rB = 22 + 42 = 20 VPQ = 15 10-9/40 [ 1/ 14 - 1/ 20 ] VPQ = 5.885 V VPQ = VP - VQ ( VQ = 5 ) So VP = VPQ + VQ = 5.85 V + 5V = 10.885 V ANSWER

**D4.6**. If we take the zero reference for potential at infinity, find the potential
at (0, 0, 2) caused by this
charge configuration in free space (a) 12 nC/m on the line 0 3C 1 = 2.5 m, z = 0;
(b) point charge of 18 nC at
(1, 2,−1); (c) 12 nC/m on the line y = 2.5, z = 0, −1.0 < x < 1.0.

**Solution:** (a) Reference
infinity
P(0, 0, 2)
L = 12 nc/m = 2.5m ; z = 0
P(0, 0, 2) r = 2 + z2
z = 2.5m Professor Aatif Saeed Preston University Islamabad Page 23 2 V
= L dL
0 40 r
2 V = L (dap +da + dzaz)
0 40 r As = 2.5 = constant
As z = 0 = constant
So we have d = 0
dz = 0 2 V = L d
0 40 r 2 V = 1210-9 (2.5) d
0 40 (3.2) ( r = 2.52 + 22 = 3.2 ) 2 V = 84.25 d
0 V = 84.25 ( ) 2
0 = 84.25 2 ( b) = 529.41 volts
Point charge Q = 18 nC P
at Q(1,2,-1)
r V = Q
40 r E Where r = Q - P
= 1,2,-3 r = 12 +22 +(-3)2
= 14

V = 1810-9 40 14 Q = 43.25 volts Professor Aatif Saeed Preston University Islamabad Page 24 (c) P(0,0,2) r y = 2.5 V = L 20 r ( r = 2.52 + 22 = 3.2 ) V = = 12 10-9 20 (3.2) 67.4 volts ANSWER

**D4.7.** A portion of a two-dimensional (Ez = 0) potential field is shown in
Figure 4.7. The grid
lines are 1 mm apart in the actual field. Determine approximate values for
E in rectangular
coordinates at: (a) a; (b) b; (c) c.

**Solution:
**Each grid line = 1mm
(a) for ‘a’ V = 106 -104 = 2 v
L = 2 grid = 2mm
vL = 1000 v/m
But E = - vL = - 1000 v/m (b) 108
r
106 y
x V = 108 -106 = 2 v
Infigure x 1.5 grid = 1.5 mm
Professor Aatif Saeed Preston University Islamabad Page 25 y 1.9 grid =
1.9 mm r = x2 + y2 r = 1.52 + 1.92 = 2.42 Vx 0.9 Vy 1.3 Vx /x = 0.9/1.5 =
600 Vy/y = 1.3/1.9 = 684 E = - (Vx /x )ax - (Vy/y )ay
= (c) -600 ax – 684 ay same as part (b)

**D4.8.** Given the potential field in cylindrical coordinates, V = 100( 0 3C 1 cos
φV)/ z2 +
1 and poi...