Source: https://b-ok.org/book/3554111/0581a3
Timestamp: 2019-04-20 19:15:21+00:00

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a) a unit vector in the direction of −M + 2N.
(5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6.
1.2. Vector A extends from the origin to (1,2,3) and vector B from the origin to (2,3,-2).
φ = 150 , and so t = sin 150 ax − cos 150 ay = 0.5(ax + 3ay ).
and where B = 12 + 32 + 22 = 14. Therefore cos θ = 5/14, so that θ = 69.1◦ .
|z| < 2; 4) the plane x = π/2, with |y| < 2, |z| < 2.
with |x| < 2, |y| < 2, |z| < 1.
y 2 sin 2x = 0. This condition is met on the plane y = 0, with |x| < 2, |z| < 2.
ambiguity exist when the dot product is used?
using the dot product, so no ambiguity.
and |Gp | = 5. Thus aG = (0.6, 0.8, 0).
(0.6, 0.8, 0) · (1, 0, 0) = 0.6. Thus θ = 53◦ .
1/3 ⇒ θ = 70.53◦ . This result (in magnitude) is the same for any two diagonal vectors.
a) the vector RM N : RM N = (−0.2, 0.1, 0.3) − (0.1, −0.2, −0.1) = (−0.3, 0.3, 0.4).
RM P = (−0.3, 0.3, 0.4) · (0.3, 0.2, 0.2) = −0.09 + 0.06 + 0.08 = 0.05.
(x2 , y2 , z2 ) and determine the magnitude of this vector.
a) must the three vectors be coplanar?
three vectors to sum to zero. Therefore, the three vectors must be coplanar.
b) If A + B + C + D = 0, are the four vectors coplanar?
Now, for example, if A and B lie in the x-y plane, C and D need not, as long as Cz + Dz = 0.
So the four vectors need not be coplanar to have a zero sum.
NE 34 E (not required).
The vector in the opposite direction to this one is also a valid answer.
G(3, −2, 4) = −ax + 3ay .
b) obtain a unit vector defining the direction of G at (3,-2,4).
1.20. If the three sides of a triangle are represented by the vectors A, B, and C, all directed counterclockwise, show that |C|2 = (A + B) · (A + B) and expand the product to obtain the law of cosines.
and B. Using this, we have C 2 = A2 + B 2 − 2AB cos α, which is the law of cosines.
D(−1, −4, 2) → D(ρ = 4.12, φ = −104.0◦ , z = 2).
a = −0.90aρ − 0.44az .
1.22. A sphere of radius a, centered at the origin, rotates about the z axis at angular velocity Ω rad/s.
The rotation direction is clockwise when one is looking in the positive z direction.
reversed. The answer is v(x, y) = Ω [−y ax + x ay ], where (x2 + y 2 + z 2 )1/2 < a.
1.23. The surfaces ρ = 3, ρ = 5, φ = 100◦ , φ = 130◦ , z = 3, and z = 4.5 define a closed surface.
and φ2 with the x axis respectively.
a) Find E at P : E = 1.10aρ + 2.21aφ .
b) Find |E| at P : |E| = 1.102 + 2.212 = 2.47.
obtain F(ρ, φ) = 5(cos φ aρ − sin φ aφ ).
−5 sin φ. Combining, we obtain F(r, θ, φ) = 5 [sin θ cos φ ar + cos θ cos φ aθ − sin φ aφ ].
1.27. The surfaces r = 2 and 4, θ = 30◦ and 50◦ , and φ = 20◦ and 60◦ identify a closed surface.
where degrees have been converted to radians.
θ about the x axis. Therefore, A can be equal to B, but not necessarily.
b) A × ax = B × ax : This is a more restrictive condition because the cross product gives a vector.
again, A can be equal to B, but not necessarily.
vectors must lie in the same plane and lie at the same angle to x; i.e., A must be equal to B.
of part c apply, and again we conclude that A must be equal to B.
1.30. Consider a problem analogous to the varying wind velocities encountered by transcontinental aircraft.
the trip start, x = 0.
over the flight range at a value of 0.4/(1 + 0.5y 2 ) at the trip start (x = 0).
time a point charge) is located at distance R from Q1 , where R >> a.
a) What is the force on the point charge before the hemispheres are assembled around Q1 ?
electric field at the Q2 location as before, and so the force acting on Q2 is the same.
force on Q2 is now zero.
axes (and therefore each side intersects an axis at ±2.
3.3. The cylindrical surface ρ = 8 cm contains the surface charge density, ρs = 5e−20|z| nC/m2 .
integrate the charge density on that surface to find the flux that leaves it.
−∞ < y < ∞, and −d/2 < z < d/2. Find D and E everywhere.
Din = ρ0 z az C/m2 (|z| < d/2), and therefore Ein = (ρ0 z/≤0 ) az V/m (|z| < d/2).
surface is a spherical shell of radius 1 mm. The enclosed charge is the result of part a.
charge density variation with r.
To obtain the correct enclosed charge, the integrand must be ρ(r) = A/r2 .
10 mm. Let ρv = 0 for 0 < r < 8 mm.
to the surface is aρ .
radial limit is ρ. Dρ is constant over the surface and can be factored outside the integral.
To find the field outside the cylinder, we apply Gauss’ law to a cylinder of radius ρ > b.
a) What power is radiated by the region lying between latitude 50◦ N and 60◦ N and longitude 12◦ W and 27◦ W?
50◦ N lattitude and 60◦ N lattitude correspond respectively to θ = 40◦ and θ = 30◦ .
spherically-symmetric, the flux density emitted by it is I = 3.86 × 1026 /(4πr2 ) ar W/m2 .
b) What is the power density on a spherical surface 93,000,000 miles from the sun in W/m2 ?
−4 nC/m2 , and ρs0 , respectively.
and so Dr = 0.
So Dr (r = 3) = 8.9 × 10−9 C/m2 .
So Dr (r = 5) = 6.4 × 10−10 C/m2 .
= −(4/9) × 10 C/m2 .
3.14. A certain light-emitting diode (LED) is centered at the origin with its surface in the xy plane.
the z axis), and r is the radial distance from the origin at which the power is detected.
ρv = 4ρ µC/m3 for 1 < ρ < 2 mm.
from which Dρ (2.4mm) = 3.9 × 10−6 µC/m2 .
3.16. An electric flux density is given by D = D0 aρ , where D0 is a given constant.
Q = ∇ · DØcenter × ∆v = 12.83 × (0.2)3 = 0.1026 Close!
freezes. Therefore the net outward flux of thermal energy through the surface is positive.
must also be positive. Answer: positive.
small volume is zero. Therefore the divergence must be zero.
in density at any point. Divergence is therefore zero.
3.19. A spherical surface of radius 3 mm is centered at P (4, 1, 5) in free space. Let D = xax C/m2 .
where ρ0 , a, and b are constants.
hemispherical surface, r = a, 0 < θ < π/2, 0 < φ < 2π.
is a much easier calculation).
Everywhere else, D = 0.
the plane, the net charge is zero.
This is equivalent to the net inward flux of D into the volume, as was found in part b.
which we evaluate at r = 4 to find ρv (r = 4) = 17.50 C/m3 .
the same as the outward flux, or again, Q = 320π C.
leads to the continuity equation, ∇ · (ρm U) = −∂ρm /∂t.
volume at the same point, as the continuity equation states.
constant closed surface, S, and explain the physical significance of the equation.
of change in total mass within the enclosed volume.
which when evaluated at r = 0.06 yields ρv (r = .06) = 1.20 mC/m3 .
b) Find ρv for r = 0.1 m: This is in the region where the second field expression is valid.
so the volume charge density is zero at 0.1 m.
3.27c) What surface charge density could be located at r = 0.08 m to cause D = 0 for r > 0.08 m?
3.28. Repeat Problem 3.8, but use ∇ · D = ρv and take an appropriate volume integral.
the right and left surfaces, and those at z = 3 and z = 2 the top and bottom surfaces.
only: The permittivity can be written as ≤(x) = ≤1 exp(α1 x), where ≤1 and α1 are constants.
where E0 is a constant.
the permittivity can be written as ≤(r) = ≤2 exp(−α2 r), where ≤2 and α2 are constants.
4.1. The value of E at P (ρ = 2, φ = 40◦ , z = 3) is given as E = 100aρ − 200aφ + 300az V/m.
sin(63.4) = 0.894. Substituting these, we obtain dW = 3.1 µJ.
as well position the two points at the same z location and the problem would not change.
direction (roles of P and Q) reversed.
easier way to obtain this result?
where we use cos2 φ + sin2 φ = 1.
4.6. An electric field in free space is given as E = x âx + 4z ây + 4y âz . Given V (1, 1, 1) = 10 V.
arc, the circle centered at the origin, from x = a to x = y = a/ 2.
where q = 1, x = a cos φ, and y = a sin φ.
4.10. A sphere of radius a carries a surface charge density of ρs0 C/m2 .
√ such here.√To evaluate the constant, C, we first look at point M , where VT = 0.
With a zero reference at r → ∞, C = 0 and therefore V (r) = 1/(r2 + a2 ).
V (r) = V0 a2 /r2 , where V0 and a are constants.
the x axis from x = a to +∞, where a > 0. Assume a zero reference at infinity.
where ρ0 and a are constants.
Then ρv = ∇ · D = 0.
b) |V |. This will be just 15.0 V.
d) |E|P : taking the magnitude of the part c result, we find |E|P = 75.0 V/m.
f) D: This is DØ = ≤0 EØ = 62.8 ax + 202 ay − 629 az pC/m2 .
Q = −2π(.6)(1)521 × 10−12 C = −1.96 nC.
where ρ0 is a constant.
e) Find the stored energy in the field by an integral of the form of Eq. (44) (not Eq. (45)).
4.25. Within the cylinder ρ = 2, 0 < z < 1, the potential is given by V = 100 + 50ρ + 150ρ sin φ V.
At P , this is ρvP = −443 pC/m3 .
first quadrant. The potential at any point in the plate is given as V = −e−x sin y.
entry is that of −E at (0, π/3), or 3/2 ax − 1/2 ay .
the field expression evaluated at the exit point, we find the direction on exit to be −ay .
4.27. Two point charges, 1 nC at (0, 0, 0.1) and −1 nC at (0, 0, −0.1), are in free space.
Taking the magnitude of the above, we find |EP | = 25.2 V/m.
what conditions does the answer agree with Eq. (34), for the potential at θa ?
This result agrees with Eq. (34) if θa (the ending point in the path) is 90◦ (the xy plane).
the field) to the xy plane; if θb < 90◦ , negative work is done since we move with the field.
4.29. A dipole having a moment p = 3ax − 5ay + 10az nC · m is located at Q(1, 2, −4) in free space.
surface on which Ez = 0 but E 6= 0?
The above becomes zero on the cone surfaces, θ = 54.7◦ and θ = 125.3◦ .
4.31. A potential field in free space is expressed as V = 20/(xyz) V.
This, multiplied by a cube volume of 1, produces an energy value of 207 pJ.
4.35. Four 0.8 nC point charges are located in free space at the corners of a square 4 cm on a side.
direction between z = 0 and 1, and x = 0 and 2.
The average current density is thus Javg = (77.4/1.46) ar = 53.0 ar A/m2 .
is reasonable to assume that J is not a function of θ or φ.
by a time-varying charge density.
since the integrals will cancel each other.
d) Show that the divergence theorem is satisfied for J and the surface specified in part b.
circular disk of radius R, centered on the z axis and located at a) z = 0; b) z = h.
R2 + h2 − h .
write a continuity equation for mass.
between the two circular faces.
is at distance z = ` from the x-y plane; the wide end (2-mm radius) lies at z = ` + 16 cm.
` is chosen such that if the cone were not truncated, its vertex would occur at the origin.
compared to its radius as measured from the z axis (ρ). This is our primary assumption.
from which we identify the resistance as R = 0.40/π = 0.128 ohms.
5.10. A large brass washer has a 2-cm inside diameter, a 5-cm outside diameter, and is 0.5 cm thick.
z axis is the axis of the washer.
c) What is the resistance between the two faces?
5.11. Two perfectly-conducting cylindrical surfaces of length l are located at ρ = 3 and ρ = 5 cm.
which is in agreement with the power density integration.
S/m. A current of 200 A dc is flowing down the tube.
The voltage drop is now V = IR1 = 200(7.38 × 10−4 = 0.147 V.
RT = R1 R2 /(R1 + R2 ) = 7.19×10−4 Ω, and the voltage drop is now V = 200RT = .144 V.
5.14. A rectangular conducting plate lies in the xy plane, occupying the region 0 < x < a, 0 < y < b.
An identical conducting plate is positioned directly above and parallel to the first, at z = d.
note that the line integral of E between the bottom and top plates must always give V0 .
Therefore E = −V0 /d az V/m.
5.15. Let V = 10(ρ + 1)z 2 cos φ V in free space.
constant. The inner conductor is charged to potential V0 , and the outer conductor is grounded.
At z = 0, we use this to find D(z = 0) = ≤0 E(z = 0) = −100≤0 x/(x2 + 4) az C/m2 .
result is 0 for all x and y.
material having radial-dependent conductivity, σ(ρ) = σ0 ρ, where σ0 is a constant.
5.19. Let V = 20x2 yz − 10z 2 V in free space.

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