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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- 3 rN|.
SOLUTION:
(a) RMN M(-1,2,1)
RMN = rN - rM
RMN = (3x- 3 ay+ 0 az)-(- 1 ax+ 2 ay+ 1 az) N(3,-3,0) rM
rN
= (3+1)ax +(- 3 - 2)ay +(0-1)az
RMN = 4ax-5ay-az
(b) RMN + RMP =? M(-1,2,1)
RMP = rP - rM
rM
= (-2,-3,-4) – (-1,2,1) P(-2,-3,-4)
= (-2+1)ax +(- 3 - 2)ay+(- 4 - 1)az rP
= - 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.
(d) aMP =?
aMP = RMP
|RMP|
RMP = - ax-5ay-5az
______________
|RMP| = (-1)^2 + (-5)^2 +(-5)^2
___
= √
aMP = - ax-5ay-5az/7.
=-0.14ax-0.7ay-0.7az
(e) |2rP- 3 rN|=?
rP = - 2ax – 3ay-4az 2rP = - 4ax-6ay-az rN = 3ax - 3ay+0az 3rN = 9ax-9ay+0az
2 rP- 3 rN = (-4ax-6ay-az) – (9ax-9ay+0az)
= - 13ax+3ay-8az
______________
|2rP- (^3) rN|= (-13)^2 + (3)^2 +(8)^2 = √ 242 |2rP- (^3) rN| = 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.
(b) aS = S
│ S │
│ S │ = (5.95)^2 +(11.9)^2 +(23.8)^2 = 35.40+141.61+556. = 733.
│ S │ = 27.
= 5.95 ax +11.9ay +23.8az /27.
= 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
(b) RAC =rC - rA
R AB ( 2 a x 3 a y 4 a z ) (6 a x a y 2 a z )
8 a x 4 a y 6 a z
RAC =rC - rA
RAC = (-3,1,5)-(6,-1,2) A
= (- 3 - 6),(1+1),(5-2) RAc rA
= - 9ax+2ay+3az C rC B
R AB R AC = ax ay az
_- 8 4 - 6
= 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 A
(b) Area =1/2│ R AB × R AC│
= 1/2│24ax - 78ay+20az│
=1/2 (24)^2 + (-78)^2 + (20)^2 B C =1/2 576+6084+ =1/2 7060
(c) Unit vector perpendicular to the plane in which triangle is located =?
Unit vector of R AB × R AC = R AB × R AC │ R AB × R AC│
= 24ax - 78ay+20az /84.
= 0.286ax+0.928ay+0.238az ANSWER
D1.5. (a) Give the rectangular coordinates of the point C(ρ = 4. 4 , φ =− 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: ρ = 4. 4 , φ =− 115 ◦ , z = 2
And x = ρ cos φ
y = ρsin φ z = z so x = 4.4cos(- 1150 ) x = - 1. y = 4.4sin(- 1150 ) y = - 3. 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) R ectangular coordinates into cylindrical coordinates
Given: x = − 3. 1 , y = 2. 6 , z = − 3
And ρ = x^2 +y^2
φ = tan-^1 y/x
z = z
Now ρ = (-3.1)^2 +(2.6)^2
φ = tan-^1 (2.6/-3.1)
Angle is in II quadrant so
φ = 180^0 - 39. 90
φ = 140^0
z = - 3
therefore cylindrical coordinates of point D are ρ = 4.05, φ = - 3.99^0 , z = - 3
D( ρ = 4.05, φ = - 3.99^0 , 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
= ( - 3.1 ax-2.6ay-3az)- ( - 1.860ax-3.99ay+2az)
= - 1.24ax+6.59ay-5az
│RCD│ = (-1.24)^2 +(6.59)^2 +(-5)^2
│RCD│ =8.36 ANSWER
D1.6. Transform to cylindrical coordinates: (a) F = 10 ax − 8 ay + 6 az at point P(10, − 8 , 6); (b)G = (2x + y)ax − (y − 4 x)ay at point Q(ρ, φ, z). (c) Give the rectangular components of the vector H = 20 aρ − 10 aφ + 3 az at P(x = 5 ,y = 2 , z = − 1).
Solution:
(a) F = 10 ax − 8 ay + 6 az
FP = F.aP = ( 10 ax − 8 ay + 6 az).aP
= 10 ax. aP - 8 ay. aP + 6 az. aP
= 10cos φ-8sinφ+0 (1)
For point P, x = 10 ,y = - 8
Φ = tan-^1 y/x
= tan-^1 (-8/10)
= - 38.6598^0
As y is – ve and x is +ve φ is in the 4th^ quadrant.hence φ calculated is correct
Cosφ=0.7808 ,sinφ= - 0.6 246
(1) FP = 10(0.7808)-8(-0.6246)
= 12.804aP
F φ = F.aφ = ( 10 ax − 8 ay + 6 az).aφ
= 10 ax. a φ − 8 ay. a φ + 6 az. a φ
= 10(-sinφ)-8(cosφ)+
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,
r = x^2 +y^2 +z^2
= - 32 +2^2 +1^2
= 3.
= cos-^1 z
x2+y2+z
= cos-^1 (1/3.742)
φ = tan-^1 y/x
φ = tan-^1 2/- 3
= - 33.69^0 +180^0
= 146.31^0
Hence C(r=3.742, =74.50^0 , φ =146.31^0 )
(b) D ( r = 5 , θ = 20 ◦, φ = − 70 ◦)
X = rsin cosφ y = rsin sinφ
= 5sin(20)cos(- 700 ) =5sin(20)sin(- 700 ) =5(0.342)(0.342) =5(0.342)(-0.939) = 0.585 = - 1.
Z = rcos
= 5cos(20^0 ) = 5(0.939) = 4.
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
│RCD│ = (3.585)^2 +(-3.607)^2 +(3.7)^2
│RCD│ =6.289 6.29 ANSWER
D2.1. A charge QA = − 20 μC is located at A( − 6 , 4 , 7), and a charge QB = 50 μC is at
B(5, 8 , − 2) in free space. If distances are given in meters, find:
(a) RAB; (b) │RAB│. Determine the vector force exerted on QA by QB if € 0 =: (c) 10-
9 / (36π) F/m; (d) 8. 854 × 10^ -^12 F/m.
Solution: (a) RAB = rB - rA
= (5-(-6))ax +(8-4)ay +(- 2 - 7)az = 11ax+4ay - 9az (b) │RAB│= (11)^2 +(4)^2 +(-9)^2
= 14.76m
(c) FAB = QAQBRAB
4 o │RAB│^3 = ( - 20 10 -^6 )(50 10 -^6 )( 11ax+4ay - 9az )/4 (10-^9 /36 )│14.7│^3 = - 9 103 10 -^6 10 -^6 109 ( 11ax+4ay - 9az ) │14.7│^3 =30.76ax+11.184ay - 25.16az mN
(d) FAB = QAQBRAB
4 o │RAB│^3 = ( - 20 10 -^6 )(50 10 -^6 )( 11ax+4ay - 9az )/4 8.85 10 -^12 │14.7│^3 = - 103 10 -^12 1012 ( 11ax+4ay - 9az) /4 8.85 │14.7│^3 =30.72ax+11.169ay - 25.13az Mn ANSWER
D2.2. A charge of − 0. 3 μC is located at A(25, − 30 , 15) (in cm), and a second charge of
0. 5 μC 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 E 0 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
4 o │ROA│^3
ROA = (0-(25))ax +(0-(- 30 ))ay +(0-15)az cm
= - 25ax+30ay - 15az
│ROA│= (-25)^2 +(30)^2 +(-15)^2
=41.83cm
EA = (-0.3 10 -^6 )( - 25ax+30ay - 15az)/4 8.85 10 -^12 │41.83 10 -^2 │^3
= - 368.55(-25ax+30ay - 15az)
EB = QBROB
4 o │ROB│^3
ROB = (0-(-10))ax +(0-8)ay +(0-12)az cm
= 10ax-8ay - 12az
│ROB│= (10)^2 +(-8)^2 +(-12)^2
=17.55cm
EB = (0.5 10 -^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
0.1 0.1 0.1 0.2 0.2 0. (a) Q = vol ρdv = 1/ x^3 y^3 z^3 dxdydz + 1/ x^3 y^3 z^3 dxdydz 0.2 0.2 0.2 0.1 0.1 0. = 1 _ 1 = 1 _ 1 = 0 0.1 0.1 0.1 0.2 0.2 0.
8x^2 │ y^2 │ z^2 │ 8x^2 │ y^2 │ z^2 │ 8 (0.03) 8 (0.03)
0.2 0.2 0.2 0.1 0.1 0.
(b) ρ^3 z^2 sin0.6 φdzdρdφ
= ρ^3 dρ z^2 dz sin0.6dφ
= (-cos0.6φ/0.6)│ (ρ^4 /4)│ (z^3 /3)│ =1.018 Mc 0 0 0
(c) e
- 2r
sin dφd dr
= e
2rdr sin d dφ
0 0 0
= (-e
= - 6.28 C 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
2 0 px 2 0 py = 5 10 -^9 0ax +0ay+4 az + 5 10 -^9 0ax +0ay+4 az 2 0 (4) 0 +0+4^2 2 0 (4) 0 +0+4^2
= 22.47az + 22.47az
= 44.939 az v/m
(b) E(PB) = PLx .apx + PLy .apy
2 0 px 2 0 py = 5 10 -^9 0ax +3ay+4 az + 5 10 -^9 0ax +3ay+4 az
2 0 (5) 0 +3^2 +4^2 2 0 (4) 0 +3^2 +4^2 = 5 10 -^9 (0.6ay+0.8az) + 5 10 -^9 .az 2 0 (5) 2 0 (4) = 10.785ay +36.850az v/m ANSWER
D2.6. Three infinite uniform sheets of charge are located in free space asfollows: 3 nC/m 2 at z = − 4, 6
nC/m 2 at z = 1, and − 8 nC/m 2 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/m 2 : E 1 = s aN = 3 10 -^9 aZ = 169.5 aZ
2 0 2 0 (ii) Electric field due to 6 nC/m 2 : E 1 = s aN = 6 10 -^9 aZ (^) = 338.8 az 2 0 2 0
(iii) Electric field due to - 8 nC/m 2 : E 1 = s aN = - 8 10 -^9 aZ = - 451.76 az
2 0 2 0
According to direction of point relative to normal
(a) E = - E 1 – E 2 – E 3 = - 56.6 az v/m
(B) E = + E 1 – E 2 – E 3 = 283 az v/m
(c) E = + E 1 + E 2 – E 3 = 961 az v/m
(d) E = + E 1 + E 2 + E 3 = 56.6 az v/m
D2.7. Find the equation of that streamline that passes through the pointP(1, 4 , − 2) in the field E = (a)
− 8 x/yax + 4 x 2 /y 2 ay; (b) 2e^5 x^ [y(5x + 1)ax + xay ].
Solution :
(a) E = ( - 8x/y )ax + ( 4x^2 /y^2 )ay P (1,4,-2 )
dy/dx = Ey/Ex
dy/dx = ( - 8x/y ) ( 4x^2 /y^2 )
dy/dx = - x/2y
2ydy = - xdx
2 ydy = - xdx
x^2 +2y^2 = c
0
D = 0 E = 60 az ANSWER
D3.3. Given the electric flux density, D = 0. 3 r 2 ar nC/m 2 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, = 25^0 , = 90^0 )
D = 0 E
E = D 0
= (0. 3 r^2 arnCm^2 ) 8. 85 10 -^12 Fm
= (0. 3 22 arnCm^2 ) 8. 85 10 -^12 Fm
= 135. 5 arVm
(b) Q =? for a sphere of radius r=
Q = ∮ 𝐷. 𝑑𝑠
ds = r^2 sindd Q = 0.3r^2 10 -^9 ∫ 2
0 ∫^ 𝑟^2 𝑠𝑖𝑛𝑑𝑑
0
= 0.3 r^2 (4r^2 ) 10 -^9
= 1.2r^4 10 -^9
= (1.2r^4 )r=4 10 -^9
= 305 nc
( c ) = Q =?
for a sphere of radius r=
= 1.2r^4 10 -^9
= (1.2r^4 )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 1/7 C at ( − 1 , 2 , − 2); (b) a uniform
line charge of π C/m at x = − 2, y = 3; (c) a uniform surface charge of 0. 1 C/m 2 on the plane y = 3 x.
Solution :
(a) since both the given charges are enclosed by the cubical volume according to the gauss's law = Q 1 + Q 2
= 0. 1 C + (17)C = 0. 243 C
(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. 4 C
( 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. 10 3 10 y = 3x 10 The length is moving up and down along z axis between z = 5 By putting y=5 and x=5/ L 1 = (5)^2 + ( 5/3)^2 = 25 + 25 9 L 1 = 5. The same length we will get on the plane formed by – ve x and – ve y axis L 2 = (-5)^2 + (- 5/3)^2 = 25 + 25 9 L 2 = 5. L = 5.270 + 5. = 10. Now this straight line is moving between z = 5 to form a plane whose area is given by 10 10 .540 = 105. 40 surface charge density s = 0. 1 C now according to gauss's law = Qenclosed
= s (area of the plane)
= 0. 1 C (105.40) = 10. 54 C 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/m 2 at r = 1 cm, and − 0 .6 mC/m 2 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 Q 1 = 0.25c
D = (Q 4 r^2 )ar D = (0. 25 10 -^6 4 (0. 5 10 -^2 )^2 ar ( as r=0.5cm and Q 1 = 0.25c )
= 796 arC
(b) Q 2 = s (area of the sphere) = 2 10 -^3 cm 4 (1 10 -^2 )^2 ( r = 1cm) = 2.513 10 -^6 c D = (Q 1 + Q 2 4 r^2 )ar D = (0. 25 10 -^6 + 2.513 10 -^6 4 (1. 5 10 -^2 )^2 ar ( r = 1.5 cm)
= 977 arC
( c ) Q 3 = s (area of the sphere)
= - 0.6 10 -^3 cm 4 (1.8 10 -^2 )^2
Dy = (4x^2 z2) = 0 y y Dz = (16x^2 yz3) = - 1728 z z
Q = Dx + Dy + Dz V
x y z = ( - 648 – 1728 ) 10 -^12 10 -^12 m^3 ( p = 10-^12 )
= - 2.38 -^21 c ANSWER
D3.8 .Determine an expression for the volume charge density associated witheach D field: (a) D
= 4 xy/z ax + 2 x 2 / z ay − 2 x^2 y/z^2 az
Solution:
D = 4 xy/z a x + 2 x 2 / z a y − 2 x^2 y/z^2 a z
divD = (4xyz-^1 ) + (2x^2 z-^1 ) - (2x^2 yz-^2 )
x y z
= 4yz-^1 + 0 - 2(-2)(x^2 yz -^3 ) = 4y + 4 x^2 y z z^3 = 4yz^2 + 4x^2 y z^3 = 4y (x^2 + z^2 ) z^3 D4.1. Given the electric field E = 1 / z^2 (8xyzax + 4 x 2 zay − 4 x 2 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 − 27 az; (c) 37ax + 67 ay.
Solution:
E = 1 z^2 (8 xyz a x + 4 x 2 z a y − 4 x 2 y a z ) v/m
dL = 2 m Q = 6-nc P (2 , − 2 , 3)
(a) aL = −6/7 a x + 3/7 a y + 2/7 a z
dL = aL. dL
= 2 m ( −6/7 a x + 3/7 a y + 2/7 a z ) = 2 10 -^6 m ( −6/7 a x + 3/7 a y + 2/7 a z )
Q = - QE.dL
Q = (- 6 10 -^9 ) 1 z^2 8 xyz a x + 4 x^2 z a y – 4 x^2 y a z .dL = - 6 10 -^9 2 10 -^6 1 z^2 8 xyz a x + 4 x^2 z a y – 4 x^2 y a z . −6/7 a x + 3/7 a y + 2/7 a z = - 12 10 -^15 1 z^2 ( 8 xyz)(-6/7) + ( 4 x^2 z )(3/7)– ( 4 x^2 y)(2/7) = - 12 10 -^15 1 z^2 -48/7 xyz + 12/7 x^2 z – 8/7 x^2 y
Putting values x = 2 , y = - 2 and z = 3
= - 12 10 -^15 1 (3)^2 (-48/7) (2)(-2)(3) + (12/7) (2)^2 (3) – (8/7 )(2)^2 (-2)
= - 12 10 -^15 1 9[ 576/7 + 144/7 + 64/7 ]
= - 4 10 -^15 1 3 [576+144+64]
= - 4/3 10 -^15 [ 112 ]
= - 448/3 10 -^15
= - 149.3 10 -^15
= - 149.3 fJ
(b) E = 1 z^2 (8 xyz a x + 4 x 2 z a y − 4 x 2 y a z ) v/m
dL = 2 m Q = 6-nc P (2 , − 2 , 3) aL = 6/7 a x - 3/7 a y - 2/7 a z
dL = aL. dL
= 2 m (6/7 a x - 3/7 a y - 2/7 a z ) = 2 10 -^6 m (6/7 a x - 3/7 a y - 2/7 a z )
Q = - QE.dL
Q = (- 6 10 -^9 ) 1 z^2 8 xyz a x + 4 x^2 z a y – 4 x^2 y a z .dL = - 6 10 -^9 2 10 -^6 1 z^2 8 xyz a x + 4 x^2 z a y – 4 x^2 y a z . 6/7 a x - 3/7 a y - 2/7 a z = - 12 10 -^15 1 z^2 ( 8 xyz)(6/7) + ( 4 x^2 z )(-3/7)– ( 4 x^2 y)(-2/7) = - 12 10 -^15 1 z^2 48/7 xyz - 12/7 x^2 z + 8/7 x^2 y
Putting values x = 2 , y = - 2 and z = 3
= - 12 10 -^15 1 (3)^2 (48/7) (2)(-2)(3) + (-12/7) (2)^2 (3) + (8/7 )(2)^2 (-2)
= - 12 10 -^15 1 9[ - 576/7 - 144/7 - 64/7 ]
W = - 4 5.dx B
Where dL = dxax+dyay+dzaz
A
= - 20X
B
= 20 j
(b) Q = 4C A (0 , 2 , 0) B (1 , 0 , 0) Path y = 2- 2x z = 0
E = 5x ax v/m
A W = - Q E.dL B A W = - 4 5x.dx B
Where dL = dxax+dyay+dzaz
A
= - 20X^2 /
B
= 10 j
( c ) E = 5 x a x + 5 yay
A W = - Q E.dL B A W = - 4 ( 5 x a x + 5 yay) .dx B
Where dL = dxax+dyay+dzaz
A
= - 20 x^2 /2 + y^2 /
B
= - 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 straight- line 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 = - Q E.dL
B (2,3,5) (2,0,5) (2,0,3) W = - 3 E.dL + E.dL + E.dL (1,3,5) (2,3,5) (2,0,5)
Where E.dL = ( yax ) .( dxax+dyay+dzaz)
= ydx
W = - 3 ydx + ydx + ydx
A 1 A 2 A 3 As X is only varing in the case of A 1 so dx = 0 for A 2 , A 3
W = - 3 ydx as y is a constant for A 1 so y = 3
A 1
= - 3 (3)dx
A 1 = - 9 x A 1 = - 9(2-1) = - 9 Joules
(b) (1,3,3) (1,0,3) (2,0,3)
W = - 3 E.dL + E.dL + E.dL (1,3,5) (1,3,3) (1,0,3)
Where E.dL = ydx
As dx 0 for the third integral only; so the first two integrals are zero.
So w = - 3 ydx ( y = constant = 0 )
= - 9(0)dx = 0 Joules ANSWER D4.4. An electric field is expressed in rectangular coordinates byE = 6 x^2 ax + 6 yay
+ 4 azV/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 = 6 x^2 ax + 6 yay + 4 az V/m
M = (2 , 6 , −1) , N = ( − 3 , − 3 , 2)
A
V = - E.dL B
