Download - Midterm HW Answers
Problem 3.1 Vector A starts at point (1,!1,!3) and ends at point (2,!1,0). Finda unit vector in the direction of A.
Solution:
A= x(2!1)+ y(!1! (!1))+ z(0! (!3)) = x+ z3,|A| =
"1+9= 3.16,
a=A|A| =
x+ z33.16
= x0.32+ z0.95.
Problem 3.3 In Cartesian coordinates, the three corners of a triangle areP1 = (0,4,4), P2 = (4,!4,4), and P3 = (2,2,!4). Find the area of the triangle.
Solution: Let B=!"P1P2 = x4! y8 andC=
!"P1P3 = x2! y2! z8 represent two sides of
the triangle. Since the magnitude of the cross product is the area of the parallelogram(see the definition of cross product in Section 3-1.4), half of this is the area of thetriangle:
A= 12 |B#C| =
12 |(x4! y8)# (x2! y2! z8)|
= 12 |x(!8)(!8)+ y(!(4)(!8))+ z(4(!2)! (!8)2)|
= 12 |x64+ y32+ z8| = 1
2
!
642+322+82 = 12$5184= 36,
where the cross product is evaluated with Eq. (3.27).
Problem 3.5 Given vectors A= x+ y2! z3, B= x2! y4, and C= y2! z4, find(a) A and a,(b) the component of B along C,(c) θAC,(d) A"""C,(e) A ·(B"""C),(f) A""" (B"""C),(g) x"""B, and(h) (A""" y) · z.
Solution:(a) From Eq. (3.4),
A=!
12+22+(!3)2 =#14,
and, from Eq. (3.5),
aA =x+ y2! z3#
14.
(b) The component of B along C (see Section 3-1.4) is given by
BcosθBC =B ·CC
=!8#20
= !1.8 .
(c) From Eq. (3.18),
θAC = cos!1A ·CAC
= cos!14+12#14#20
= cos!116#280
= 17.0$.
(d) From Eq. (3.27),
A"C= x(2(!4)! (!3)2)+ y((!3)0!1(!4))+ z(1(2)!2(0)) = !x2+ y4+ z2.
(e) From Eq. (3.27) and Eq. (3.21),
A · (B"C) = A · (x16+ y8+ z4) = 1(16)+2(8)+(!3)4= 20.
Eq. (3.30) could also have been used in the solution. Also, Eq. (3.29) could be usedin conjunction with the result of part (d).(f) By repeated application of Eq. (3.27),
A" (B"C) = A" (x16+ y8+ z4) = x32! y52! z24.
Eq. (3.33) could also have been used.(g) From Eq. (3.27),
x"B= !z4.
(h) From Eq. (3.27) and Eq. (3.21),
(A! y) · z= (x3+ z) · z= 1.
Eq. (3.29) and Eq. (3.25) could also have been used in the solution.
Problem 3.7 Given A= x(x+2y)! y(y+3z)+ z(3x! y), determine a unit vectorparallel to A at point P= (1,!1,2).
Solution: The unit vector parallel to A = x(x+2y)! y(y+3z) + z(3x! y) at thepoint P= (1,!1,2) is
A(1,!1,2)|A(1,!1,2)|
=!x! y5+ z4
!
(!1)2+(!5)2+42=
!x! y5+ z4"42
#!x0.15! y0.77+ z0.62.
Problem 3.11 Find a unit vector parallel to either direction of the line described by
2x+ z= 4.
Solution: First, we find any two points on the given line. Since the line equationis not a function of y, the given line is in a plane parallel to the x–z plane. Forconvenience, we choose the x–z plane with y= 0.For x= 0, z= 4. Hence, point P is at (0,0,4).For z= 0, x= 2. Hence, point Q is at (2,0,0).Vector A from P to Q is:
A= x(2!0)+ y(0!0)+ z(0!4) = x2! z4,
a=A|A| =
x2! z4"20
.
Problem 3.15 A certain plane is described by
2x+3y+4z= 16.
Find the unit vector normal to the surface in the direction away from the origin.
Solution: Procedure:
1. Use the equation for the given plane to find three points, P1, P2 and P3 on theplane.
2. Find vector A from P1 to P2 and vector B from P1 to P3.
3. Cross product of A and B gives a vector C orthogonal to A and B, and henceto the plane.
4. Check direction of c.
Steps:
1. Choose the following three points:
P1 at (0,0,4),P2 at (8,0,0),P3 at (0, 163 ,0).
2. Vector A from P1 to P2
A= x(8!0)+ y(0!0)+ z(0!4) = x8! z4
Vector B from P1 to P3
B= x(0!0)+ y!
163!0
"
+ z(0!4) = y 163! z4
3.
C= A"""B= x(AyBz!AzBy)+ y(AzBx!AxBz)+ z(AxBy!AyBx)
= x!
0 · (!4)! (!4) ·163
"
+ y ((!4) ·0!8 · (!4))+ z!
8 ·163!0 ·0
"
= x 643
+ y32+ z 1283
Verify that C is orthogonal to A and B
A ·C=
!
8 ·643
"
+(32 ·0)+
!
1283
· (!4)"
=5123
!5123
= 0
B ·C=
!
0 ·643
"
+
!
32 ·163
"
+
!
1283
· (!4)"
=5123
!5123
= 0
4. C= x 643 + y32+ z 1283
c=C|C| =
x 643 + y32+ z 1283#
$ 643%2
+322+$ 1283
%2= x0.37+ y0.56+ z0.74.
c points away from the origin as desired.
Problem 4.1 A 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 ρv = xy2e!2z (mC/m3).
Solution: For the cube shown in Fig. P4.1, application of Eq. (4.5) gives
Q=!
V
ρv dV =! 2
x=0
! 2
y=0
! 2
z=0xy2e!2z dx dy dz
=
"
!112
x2y3e!2z#
$
$
$
$
$
2
x=0
$
$
$
$
$
2
y=0
$
$
$
$
$
2
z=0
=83(1! e!4) = 2.62 mC.
2 m
0
2 m
2 m y
z
x
Figure P4.1: Cube of Problem 4.1.
Problem 4.3 Find the total charge contained in a cone defined by R ! 2 m and0 ! θ ! π/4, given that ρv = 10R2 cos2 θ (mC/m3).
Solution: For the cone of Fig. P4.3, application of Eq. (4.5) gives
Q=! 2π
φ=0
! π/4
θ=0
! 2
R=010R2 cos2 θ R2 sinθ dR dθ dφ
=
"
"23R5φ cos3 θ
#
$
$
$
$
$
2
R=0
$
$
$
$
$
π/4
θ=0
$
$
$
$
$
2π
φ=0
=128π
3
%
&1"
'#2
2
(3)
* = 86.65 (mC).
2 m
0y
z
x
π/4
Figure P4.3: Cone of Problem 4.3.
Problem 4.7 If J= R5/R (A/m2), find I through the surface R= 5 m.
Solution: Using Eq. (4.12), we have
I =!
SJ ·ds=
! 2π
φ=0
! π
θ=0
"
R5R
#
· (RR2 sinθ dθ dφ)
= !5Rφ cosθ
$
$
$
$
$
R=5
$
$
$
$
$
π
θ=0
$
$
$
$
$
2π
φ=0
= 100π = 314.2 (A).
Problem 4.11 A square with sides of 2 m has a charge of 40 µC at each of its fourcorners. Determine the electric field at a point 5 m above the center of the square.
Solution: The distance |R| between any of the charges and point P is
|R| =!
12+12+52 =!27.
E=Q4πε0
"
R1|R|3 +
R2|R|3 +
R3|R|3 +
R4|R|3
#
=Q4πε0
"
"x" y+ z5(27)3/2
+x" y+ z5(27)3/2
+"x+ y+ z5
(27)3/2+x+ y+ z5(27)3/2
#
= z 5Q(27)3/2πε0
= z 5#40 µC(27)3/2πε0
=1.42πε0
#10"6 (V/m)= z51.2 (kV/m).
R3
R2
z
P(0,0,5)
y
x
R1R4
Q1(1,1,0)
Q2(-1,1,0)
Q3(-1,-1,0)
Q4(1,-1,0)
Figure P4.11: Square with charges at the corners.
Problem 4.19 Three infinite lines of charge, all parallel to the z-axis, are located atthe three corners of the kite-shaped arrangement shown in Fig. P4.19. If the two righttriangles are symmetrical and of equal corresponding sides, show that the electricfield is zero at the origin.
y
x
!2!l
!l !l
Figure P4.19: Kite-shaped arrangment of line chargesfor Problem 4.19.
Solution: The field due to an infinite line of charge is given by Eq. (4.33). In thepresent case, the total E at the origin is
E= E1+E2+E3.
The components of E1 and E2 along x cancel and their components along !y add.Also, E3 is along y because the line charge on the y-axis is negative. Hence,
E= !y 2ρl cosθ2πε0R1
+ y 2ρl2πε0R2
.
But cosθ = R1/R2. Hence,
E= !y ρlπε0R1
R1R2
+ y ρlπε0R2
= 0.
Problem 4.23 Repeat Problem 4.22 for D= xxy3z3 (C/m2).
Solution:(a) From Eq. (4.26), ρv = ∇ ·D=
∂∂x
(xy3z3) = y3z3.(b) Total charge Q is given by Eq. (4.27):
Q=!
V
∇ ·D dV =! 2
z=0
! 2
y=0
! 2
x=0y3z3 dx dy dz=
xy4z4
16
"
"
"
"
"
2
x=0
"
"
"
"
"
2
y=0
"
"
"
"
"
2
z=0
= 32 C.
(c) Using Gauss’ law we have
!
!
SD ·ds= Ffront+Fback+Fright+Fleft+Ftop+Fbottom.
Note thatD= xDx, so only Ffront and Fback (integration over z surfaces) will contributeto the integral.
Ffront =! 2
z=0
! 2
y=0(xxy3z3)
"
"
"
"
"
x=2
· (x dy dz)
=! 2
z=0
! 2
y=0xy3z3
"
"
"
"
"
x=2
dy dz=
#
$2%
y4z4
16
&
"
"
"
"
"
2
y=0
'
(
"
"
"
"
"
2
z=0
= 32 ,
Fback =! 2
z=0
! 2
y=0(xxy3z3)
"
"
"
"
"
x=0
· (!x dy dz) = !! 2
z=0
! 2
y=0xy3z3
"
"
"
"
"
x=0
dy dz= 0.
Thus Q= !
!
D ·ds= 32+0+0+0+0+0= 32 C.