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Unformatted text preview: 9 Consider a line charge parallel to the x axis, and let the hue charge density be pg(r) = 6 X
’ 10—8 C/rn. The line charge passes through the point z = 2 on the z axis. The surface So Is \h/ﬁyﬂp a sphere centered at the origin and, having a radius of 5 m. Calculate the ﬂux over So of the
h electric ﬁeld produced by the line charge. (2) Within an inﬁnitely long cylinder having a radius of 3 m, the charge density 18 given by
Q3 ‘ pvrr)— = 10 ‘°exp( p)
NW Using Gauss’s law, 1 , (a) Find the electric ﬁeld at a ﬁeld point having cylindrical coordinates of (p,d>,z), where p
is less than the radius of the cylinder. (b) Find the electric ﬁeld at a ﬁeld point having cylindrical coordinates of (p, 4);), where p
is greater than the radius of the cylinder. system so that the bottom of the cylindrical shell is in the xy plane and the axis of the shell lies
on the z axis, the surface charge density on the surface p= 2 cm is p, = 2.1 X 10—8 C/rnz.
(a) Calculate the electric ﬁeld at the ﬁeld point (x,y,z)= (0,0,15 cm) using Coulomb’s law ;’
for a surface charge. ,
(b) Consider the charge distribution on the cylindrical shell to consist of a number of hori— 5 zontal rings of line charge, as shown ' The equivalent line charge density of I
each ring is given by ' I J‘ ® A hollow cylindrical shell has a radius of 2 cm and a length of 25 cm. Choosing a coordinate P: = p,dz' = 2.1 x 10‘de C/m the contribution of this ring to the electric ﬁeld at any
arbitrary ﬁeld point on the z axis is _ ’ . 48 ‘r , ' t
dE(0,0,z) = (z z)z x 2.1 x 10 dz , (0.02) . 2 /
«2&0 (0.02)2 + (z— 5)] Verify your answer to part a by integrating this expression over all of the rings On the
cylindrical shell (as 2’ varies from O to 25 cm). @ The Surface deﬁned by
_ ‘ . z = o
1 < p < 3 O < 4) < "11' , , 43““ contains a uniform Surface charge density of 5 X‘ '10—9 'C/tn3. The curve deﬁned by the edge
of this surface contains a uniform line charge density of p50 C/m. The sum of the total Surface
charge and total line charge is equal to zero. ' ' I .  A ‘ "(11% Calculate the value of Peo
$.09) Calculate the electric ﬁeld at the ﬁeld point (x,y,z) = (0,"0,0)‘. . , C/mz. Find the electric ﬁeld at the ﬁeld points (mm) = (1,1,7) and (x7352) = (LL—10). “MK? :The surface 2 ="3x'+ 2y contains a uniform surface charge density of p5 = —3 X 10"9 Q ‘ Withinianregion deﬁned by 0.15 S 2, S +0.15 m, the volume charge" density is given by
‘X pv(r) = 10'3z2 C/m3 \W ' . . >
\WJ‘ Usmg Gauss’s law, ﬁnd an expressmn for the electric ﬁeld in each of the following three
'7 . regions: _z > 0.15 m; ‘0.15 Sz S 0.15 tn; and z < 0.15 m. i I  I A point charge of 4 X 10'8 C is located at the origin. A lineycharge characterized by pg =
(\P . 2 X 1016 C is oriented parallel to the z axis. Theline charge is 2th long, and the center of .1
V“ j the line charge is initially located at (p,¢:,z)i= (4,0,0). The line charge is then moved so that
3 the center of the line charge travels, in the direction of increasing (1), along a circular arc} "‘  having a radius of 4 111 until the center of the line charge arrives at (p,¢,z) = (4,11/2,0). While: _‘ the line charge is moved, the orientation of the line charge is always parallel to the z axis ‘ (a) Calculate the amount of work done in moving the line charge. ‘ 4 . 3 g (b) Calculate the amount of work done in moving the line charge if the line charge is unuallyi
parallel to the y axis and remains parallel to the y axis as it moves. , l \éhﬁ Ckscildispheﬁrehavmgaradms of 01th iscentered at (x,y,z) =(0,0,5)Thecharge den51ty i in the sphere is given by pv = 3.16 X 10—11 C/m3. A point charge of 1.2 X 10—12 C is located‘._f;; at the origin. Calculate the amount of work done in moving the solid sphere vertically doWni
Tl ward along the z axis until the distance between the center of the sphere and the point charge]:E
is 2.5 tn. (Hint: Consider the amount of work done if the solid sphere remains stationary53,}
5 while the point charge moves upward along the z axis from z = 0 to z = 2.5 m.)v_ . .,  . @ lewo Circular dlSkS, eachhaiﬁngradu "of 1 m, are orietitEdwpetpendicular o the x axis. One
X  disk is centered at x = 5 m on the positive x axis and has a uniform surface charge density
of p, C/ms. The ‘other disk is centered at x = 5 m on the negative x axis and also has a i\ ,
N® : uniform surface charge density of ps C/mz. A charged particle having a mass of 10"3 kg and
a charge of 10'4 C slides along the x axis between the two charged disks. At t = 0 s, the
b3 particle is located at x = 0 and is traveling in the positive x direction with a velocity of
  ‘ 10.0 m/‘s. At subsequent times, the particle oscillates between x = 2.5 and x = 2.5 m. (a) Calculate the surface charge density p: of each disk. [I—Iint: Assume thatthe total energy
of the particle is conserved as it oscillates along the x axis—i.e., the amount of work
required for the particle to move between any two points on the x axis is provided at the
expense of the particle’s kinetic energy, (1/2)mv2.] , V (b) Find an expression for the particle’s velocity as a function of the particle’s x coordinate. ‘ Q'O‘i' («l/{3’ 12060 ' ' quﬁm
, . a r I i .“ $31K “’2 i“ Willi“ t/a uh‘lgh'u (WW Maul
fetid/EV 3' am (9,929+ (ionsider a line charge along the z axis extending from z = a to z = b (where a < b). The 1 L;
line charge density is pg C/m for a < z < b and is zero everywhere else 'on the z axis. Find
an expression for the electric potential V(r) at a ﬁeld point having general coordinates of p I:
d), and z, and ﬁnd the corresponding electric ﬁeld. , 1
l
l
i
l
I
1 .\/ @ X—ﬁekcﬁarge having a uniform line charge density Of Dr C/m ‘5 011,6“th along the 2 ms and extends from 2 '= —L to z = +L.
(a) Show that the potential produced by this charge. distribution can be written as . / r
. / V( Cb z)=1n ———L — z + n \\ '
\‘\\\__ p ’471'50 "‘ “L — Z + 1'2 ' / 1 _ where r1 is the distance from the top of the charge distribution (at z = L) to the ﬁeld 1
point at (p,cb,z) and r2 is the distance from the bottom of the charge distribution (at i z = L) to the ﬁeld point. (b) Show that l
7% '— r§ = 4Lz  l (c) Using the results of parts a and b, show that the potential found in part a can also be 1
written as 1
W = 4260111 {pug—3%} . . Ilintlese  \ ' i .. V I ' , I L—z+r1 (r1+r2)—2L Lz+r1 (r1+ r75+2L 1n —— X———— =1n ————— —1n ———— Lz+r2 (r1+r2)+2L Lz,+r2 (r1+r2)—2L
and show that the left hand side of this equation is equal to zero [10].] (d) Using the result of part 0, show that the equipotential surfaces for this charge distribution
have the form of a collection of ellipsoidal surfaces described by the equation ' =1 thfDN
uLI "n where a2 and b2 are each constant for a given equipo‘tential surface. Consider a line charge of 3.1 X 10—8 cm: along the z axis. The line charge extends from H
z = 2 m to z= +2 111. As shown' 1n the previous problem, the 300 V equipotential surface
can be written in the form of ‘ 0W4 Find the numerical values of the constants a2 and b2. =1 aEliUNI
“Tel Nu @ A spherical surfaCe with a radius of 2 In is an equipotential surface at a potential of 0 V
The center of the spherical surface is at the origin. A point charge of Q= 2 X 10'8 C is ' V located on the positive z axis at z = 6 m. . (l0 (a) Find the location (on the z axis between z— — +2 and z = —2) and electrical charge (in Q
Coulombs) of an image point charge (Q) such that the absolute potential at each of the two points (x,y,z)— = (0,0 +2) and (x,y,z)— = (O, 0, +2) is equal to zero. ‘
(b) Shdw that the absolute potential due to Q and Q at any point on the spherical surface r — 2 is zero. (Hint: Use the law of cosines to ﬁnd the distance from a point on the a sphere to each of the two charges, Q and Qi.) ' "3"" (c) Using the given point charge Q and the calculated image charge (Q,), calculate the potential and the electricmﬁeld at the point (x,y,z)— = (5,7,8). \@ In the. region between two concentric cylinders (p = 0.1 m and'p ¥ '0'.4 m),the charge density 3< 'is given by
W p = 10‘8 £2— + 2p C/m3
 v 0.1 On the inner cylinder (p— 0.1), the absolute potential is 500 V. On the outer cylinder
0.4 m),E = 250 Win. Calculate the electric ﬁeld at a point midway between the inner outer cylinders. @W The electric polanga'tiouriuvec‘tor within a cube of Plexiglas 1s given by
P(r) = 3x252 and the interior of the cube 15 described by X\W :14 _ ' . O<x<O.5‘m. «@9/“~_ 
(\ OSySO.5m
O<z<0.5m (a) Find the bound surface charge on each of the six faces of the cube. '
(b) Find the total bound volume charge within the cube and show that the total bound charge (c) 20:22:61“ a cube inside the Plexiglas and let the interior of this inner cube be deﬁned by
~ * ' .  , ' 0.1 < x S O. 4 m A
0.1 S y .<_ 0.4 m
0.1 s z s 0.4 m Evaluate. 56F  ds over the surface of this inner cube. .' 7 A line charge ofl'vpe C/m per meter coincides with the z axis. The line charge is surrounded
by a cylindrical region of free space (e = e0) for O < 9 Sa. The region of free space is surrounded by a cylindrical shell of dielectric material (6. = 360) for a < p S b. The region
b < p is, again, free space. Find the values of E(r), P(r), and D(r) in all three regions. In of" addition, ﬁnd the bound surface charge density p5,, on the surfaces p = a and p = b. 0V V'EV‘WS Thai “Ne; 1,1an MM Jae/~52, n; legco. [email protected] mij 0%
697 r A cylindricalmetal conductor having a radius of 2 cm is surrounded by a lcm—thick di
electric (material 1) characterized by 6,1 = 2. Material 1 is surrounded by another dielectric
(material 2), which is 3 cm thick and characterized by 6,2 = 2.5. Material 2 is surrounded
; by a lmmthick metallic conductor. The inner conductor is grounded (V = O), and the outer
conductor is maintained at 50 V. Find the potential 'V(r) and the electric ﬁeld E(r) in‘the
region between the two conductors. , ' A solid metals here with" "A" radius of 1cm has an unknov'vn uniform free surface charge
density of ps. The sphere is surrounded by a region of free space extending from r =
1 cm to r = 1.5 cm. The region of free space is, in turn, surrounded by an insulator hav
ing an: unknown relative dielectric constant (6,). The insulator extends from r = 1.5 cm to
r.= 4 cm. The magnitude of the electric ﬂux density at a point having spherical coordi
n’ates of (r,e,¢) = (2 cm, 10°, 45°) is 3750 x 10—3/1: C/mz. The magnitude of the elec
tric polarization vector at a point having spherical coordinates of (130,43) = (4 cm, 20°, . 30°) is 468.75 x 104/7: C/mz. (a) Calculate the numerical values of p, and er .
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 Spring '10
 Ferguson
 Electromagnet

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