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Unformatted text preview: Adam S. Bolton February 15, 2002
[email protected] MIT 8.02 Spring 2002
Assignment #1 Solutions Problem 1.1 Relative strengths of gravitational and electrostatic forces. Q Q (9 ®  
 d 
The dust grains have diameter 50 am, and thus have a radius a : 25 nm : 2.5 X 10‘5 m. They have mass density ,0 = 2.5 gm/cm3 : 2.5 X 103 kg/m3 and charge Q = —ne (6 being
the magnitude of the electron’s charge). The mass of each grain is then 4
m : §7T03p E 1.6 X10_1U kg .
The gravitational attraction is
F _ Gm2
G — d2
The electrostatic repulsion is
F 10.262
C _ 47T60d2 There would be no net force if FC : F0, i.e.
n. : 47r600? E 0.09 <1 .
Thus a mere single extra electron on each grain would prevent the grains from colliding. For comparison, each grain contains m/mp : (1.6 X 10‘1°)/(1.67 X 10—27) a 1017 nucleons
(protons + neutrons). For a neutral grain, we have one electron for each proton, or one for every two nucleons, if # protons = # neutrons. Thus the total number of electrons is
E 5 X 1015. MIT 8.02 Spring 2002 — Assignment #1 Solutions 2 Problem 1.2
Electric ﬁeld along the line passing through two point charges. (Note: In these solutions, as in the text, we will always denote vector quantities such as the electric ﬁeld by boldface font: E. In lectures, and probably in yogir own handwritten worh,
vectors are more commonly written with an arrow over the top: Q1: +3uC Q2: 7 uC
+ + ex
0 0.4m (a) Let us ﬁrst determine the electric ﬁeld due to Q1 alone. From the expression for the
electric ﬁeld due to a point charge, 1 Q1
E —— ——“ .
1 471'60 rgr If we are at position a: on the xaxis, r : wit, r : Vac? : lxl, and r : r/r : and E1 2 ——.’L' Note that E1 is in the :3: direction for a: > 0 and in the —:i‘: direction for a: < 0, as we would
expect, since the charge is located at a: : 0. Similarly, _ Q2 A E _ —
2 47T60 la: — 0.43x (variables always measured in SI units unless otherwise noted). Note that E2 changes direc
tion depending on whether we are to the left or right of the charge at a: = 0.4 m. The total
E is the sum of E1 and E2: 10‘6 a: x—OA
3 —7 " .
x3 x—0.43 3’ E :
471'60 This expression is good for —00 < a: < 00, but it is worthwhile writing it out as follows: 10—6 (—3/372 + 7/(33 — 0.4)2) it“: —00 < a: < 0
— — (+3/a:2 + 7/(3: — 0.4)2)é‘: 0 < a; < 0.4
(+3/3727/(x—0.4)2)§: 0.4<a: < oo . E _
471'60 Note the sign changes as one passes over the positions of the two charges and the ﬁeld due
to that charge reverses. MI T 8.02 Spring 2002 — Assignment #1 Solutions 3 [J (b) For a: < 0, f(x):_w—2+W (See graph at right.) x (meters)
This function has a zero at a: = —0.758 m, and a maximum at a: = —1.225 m, with a value
there of 0.652 m‘2. As we move toward a: : 0, f($) blows up to —00, because of the positive
charge at a: : 0. As we move toward a: : —oo, f($) goes asymptotically to +4/x2, which simply means that far enough away >> 0.4) the ﬁeld looks like that due to a point charge
of charge Q1 + Q2 : —4 pic. (0) There are no other zeros of except at a: = —0.758 m. A glance at our expression
for above clearly shows that there can be no zeros in the range 0 < a: < 0.4 (sum of
two positive terms can never be zero). You might think that we could get a zero in the
range 0.4 < a: < 00, but the negative charge (which is of greater magnitude than the positive
charge) is closer throughout this range, and the electric ﬁeld is therefore always in the —3:
direction. Solving for the zeros of the expression for in the 0.4 < a: < 00 range will
give zeros that are outside of that range and therefore unphysical. MI T 8. 02 Spring 2002 — Assignment #1 Solutions 4 Problem 1.3
Continuous charge distribution. (Giancoli 2149.) y
A DP}: We ﬁrst ﬁnd the electric ﬁeld dB at point 0 from a portion of the arc of length dl and
located as shown in the diagram. This little element carries charge dq : A (ii. If dl subtends
an angle d6 (see diagram), then we have dl : Rd6. Now, dq
dE = :3
47reUR2 ’
where if" is a unit vector pointing from the little element towards 0. If the little piece of
arc is located at an angular position 6 as shown in the diagram, then 1" = —.i‘: cos 6 — y" sin 6.
Thus,
A
dE: —.i‘:cos6d6— “sin6d6
47reuR [ y ] To get the total E, we integrate dE from 6 : —60 to 6 : 6U, noting that so 9
f cos6d6=sin6_°9 =28iﬂ90 ;
90 0 so 9
[a sin6d6:—cos6_°6,0 :0 .
_0 Thus (The astute problem solver would have concluded from the outset, based on the symmetry of the system, that E at 0 could not conceivably have any ycomponent, and would only
have bothered to calculate Em eXplicitly.) MI T 8. 02 Spring 2002 — Assignment #1 Solutions 5 Problem 1.4
E—ﬁeid of a uniformly charged disk. This problem requires an expression for the electric
ﬁeld on the axis of a ring of radius r carrying a charge
q (see Giancoli Example 219): E — —1 —qz 2
mg _ 47T60 (22 + r2)”2 ‘ The problem at hand is a disk, not a ring,
but we may break the disk up into many
rings and add up the ﬁeld due to each, Z
using the above expression for Bring. The
area of the shaded region in the diagram
at right is 271"." dr, and thus its charge is
0(271'?‘ dr) : (Q/‘ITREXZ’ITT dr). Using our
expression above, the electric ﬁeld dE due
to the shaded area is 1
dE=2 d“ 2* Z Q2 r dr
47T60 (22 + r2)3!’2 27reuR2 (22 + N)”2 (a) R A Q2 R rdr A Q2 —1
E : E : :
[d Z27T60R2 f0 (22 + r2)”2 Z27T60R2 (22 + r2)1!’2
A Q2 1 1
E : —
(Z) ZZ'IFEDR2 z 22 + R2 The inte ral above is easily done usin the substitution 8 : V22 + r2, whereby r dr : 8 d3.
3 g
This leaves us with the simple integral f 3‘2 d3.) 0 (b) For 2 > 0, we can write this as _:2 _Z/_R
(ca/41mm) (2) [1 1+ (z/Rr We use this expression for the plot on the following page. MI T 8.02 Spring 2002 — Assignment #1 Solutions 6 ’) E (in units of 12/475805? 0 0.5 1 1.5 2 2.5 3 3.5 4
z (in units of R (c) We can understand the shape of this curve for small and large 2 by using a Taylor series
expansion. From Giancoli Appendix A—3, Let’s expand (1 + a)" about u : 0 using this formalism. First, %(1+ u)" : n(1+ u)“'1 , so that (1+u)" = (1+u)”lu=0+n(1+u)"'llu=0(u—U)+
: 1+nu+ This is the binomial expansion of Giancoli Appendix A2. It is a good approximation as
long as u is << 1. For case (i), 22 << R2, we have Q
603,2 z/R
1+ (z/Rr E 2"
(Z) ZZTF but (1+(z/R)2)‘1’l2 21—(1/2)(z/R)2 from above, so Ea = 22.6.32 [1 — (z/R) (1 — <1/2)<z/R)2)l 22 << R2 (leading term only.) MI T 8.02 Spring 2002 — Assignment #1 Solutions 7 Note if we had kept the next term, we would have an initial slope of —2 near 2 : 0, in
keeping with our plot above. For case (ii), 22 >> R2, we have 1
(R/z)2+1 but (1+(R/z)2)‘1’l2 21—(1/2)(R/z)2 for (R/z)2 <1, and E(z) 2 zgﬂgRE [1 — (1 — (1/2)(R/z)2)]
E(z) 2 23%;, .32 >> R2 ((1) Clearly the above case (ii) expression looks like Coulomb’s law for a point charge Q. (e) If we are very close to the disk (2 << R), it looks like an inﬁnite plane with surface charge
density a. The ﬁeld due to an inﬁnite plane has the same magnitude above and below the
plane, but with opposite directions (see diagram below). Applying Gauss’s law to the pillbox
shown below, we have fEdA : (send/en
EA+EA = (IA/60 2} Ezo/Zeu . (We grudgingly adopt the convenient but somewhat abstract notation dA, to be consistent
with Giancoli. We would prefer to use the more intuitive notation ﬁdS.) Since a : Q/‘rng, this expression is exactly as in c(i) above. Ml T 8. 02 Spring 2002 — Assignment #1 Solutions 8 Problem 1.5
Electric Dipole. (Giancoli 2165.) Q +Q
—l/2 l/2 7" Let the charges be located on the xaxis, with the positive charge (+Q) at a: : l/2 and the
negative charge (—62) at a: : —l/2, as shown in the diagram above. From the expression for
the electric ﬁeld due to a point charge, the electric ﬁeld on the xaxis at a: : r (> 0) due to
the positive charge is 1 (+69) E :——*
+ 471'60 (r — l/2)2x and the electric ﬁeld there due to the negative charge is _ 1 (—Q) A
E' ‘ Fem +s/2)2"” The total electric ﬁeld is 1 1
E:E++E_— Q 33. _ 47T60T2 (1 — vat)2 _ (1 +0202 Since l/2r << 1, we may to a good approximation expand the two terms in square brackets
using the binomial series expansion (as in the previous problem), and retain only the ﬁrst
order term in l/2r: (1 :l: l/Zr)_2 2 12Fl/r (for l/Zr << 1) . This gives us Ewe _ 47r60r2 269.! A 2p
.1: = i: .
47reur3 47r60r3 [(1 + 5/?)  (1  5/0133“ =
The magnitude checks out with the problem statement. E points in the positive w—direction,
as we would expect, since the positive charge is a bit closer than the negative. Notice that the net Eﬁeld is proportional to 1 /r3 whereas the Eﬁeld due to each charge separately falls
off as 1/r2. MIT 8. 02 Spring 2002 — Assignment #1 Soiutions 9 Problem 1.6 Gauss ’s Iaw and the superposition principie. Let’s choose the zaxis perpendicular to the slab and sheet, with z : 0 in the middle of the
sheet: 7
4. We will ﬁnd diﬁerent expressions for E in each of the three regions shown: Region I : z > D/2
Region II : —D/2 < z < D/2
Region III : z < —D/2 . The best way to approach this problem is to make use of the superposition principle and
symmetry arguments, i.e., to calculate the electric ﬁeld Ewe»E of the sheet of charge m,
then to calculate the electric ﬁeld Eslab of the slab of charge m, and then to add vectorially
the results. From above, we have 0. Region I : Esheet : +§2 (independent of z)
0
Region II & III : Esheet = —12 .
260
What about Eslab? Consider the fol— area A E CIA
lowing Gaussian pillbOX: the top is a “I, 1 distance 2 > D/2 above the z : 0
plane, and the bottom is the some
distance below the z : 0 plane. By
symmetry, the electric ﬁeld on the
top of the pillbox has exactly the
same magnitude, but is oppositely di
rected from the E ﬁeld on the bottom
of the pillbox. The charge Q enclosed
by the pillbox is pDA, so (in
{E  dA : Qencl/EU El, >Df2 ant—N ——2v MI T 8.02 Spring 2002 — Assignment #1 Soiutions 10 becomes ZEA : pDA/eu, or E : pD/Zeu, independent of z. was A El am
That gives us the ﬁeld outside the \§\ slab, but what about inside? Con— \\
sider a pillbox similar to the previ W $Z<Df2
ous, except now 2 < 13/2. The total __l__ _________________ __ ‘ '25:
height of this pillbox is 22, so that '
the charge contained inside is now p(22)A, and E dA 56 E  dA = owl/en => ZEA : 2sz/60 . Notice that again we used a symmetry argument by carefully choosing the pillbox to have
top and bottom the some distance from 2 : 0. Thus inside the slab, E : pz/eu. This
depends on 2, as it should! For the slab, then, D
Region I : E3131, 2 +p—23
260
Region II : Eslab 2+gz
50
D
Region III : Eslab = 49—2 .
260
In summary, we have
D
(a) RegionI (z > D/Z) : E = $2
0
(0) Region 11 (—D/2 < z < D/2) : E = [g — i 2
60 260
D
(b) Region 111 (z < —D/2) : E: —M2 . MI T 8. 02 Spring 2002 — Assignment #1 Solutions 11 (d) It is instructive to plot Esheet and Eslab separately (remembering that o < 0). 
Gf280 :(Eshcct z _D' 0 Z :  Iola’Ze0
 (Eslab z: pDJQEO _D :  0: 3“ Z
—pD!280 Together, they look like this (assuming ,0!) > cr):
EZI
_D (pD+G)J’280
I 0 } Z  (pD+G)!2£0 Note that the ﬁeld across this and ﬂ sheet of charge is discontinuous, with a jump of
magnitude 0/60. Problem 1.7
Two spherical charged shells. (Giancoli 2221.) From the symmetry of the system, we may con
clude that the electric ﬁeld is entirely in the
radial direction, and is a function of 1' alone.
We take as a Gaussian surface a spherical shell
concentric with the charged shells, and with
radius 1' in the region within which we wish to
determine the electric ﬁeld. Gauss’s law gives fEdA : Qencl/Eu
Qencl/EU  2} 47rr2E MIT 8.02 Spring 2002 — Assignment #1 Squtions 12
(a) In the region where 1' < r1, Qeml = U, and thus E = 0 there.
(b) In the region where r1 < r < r2, Qencl : 4?TT%0'1, so there
E : 0—1(E)2 :3 .
60 r
(c) In the region where r > T2, Qeml : 47T(r§01 + r302), giving
E : (rfol + r302) 13
60T2
(d) We will have E : 0 for r > m if rfol : —r§og. This amounts to having equal and opposite charges on the two shells. (e) E : 0 for r1 < r < 12 is only possible if 01 : 0, regardless of the value of 02. Problem 1.8
Two concentric charged cylinders. (Giancoli 2229.) For L >> R1,R2, we may model the system as be
ing of inﬁnite length to a good approximation. We
then conclude from the symmetry of the system that
the electric ﬁeld is directed radially outward, perpen
dicular to the axis of the cylinders, and is a function
only of the perpendicular distance r from the axis. To
apply Gauss’s law, we consider a cylindrical surface
coaxial with the charged shells and of length It << L,
with radius r in the region in which we wish to de
termine the electric ﬁeld (see diagram at right). E is
(approximately) perpendicular to the endcaps, so the
only contribution to the ﬂux integral comes from the
sides of the Gaussian cylinder. Gauss’s law gives 56 E  dA = Qend/eu => 2mm 2 (game) .
(a) For T < R1, Qend : 0 and we conclude that E : 0 in this region. (b) For R1 < r < R2, we have Qenc1 = +Qh/L, and Gauss’s law gives us Q E :
(T) 27T60L'!‘ T’: MI T 8.02 Spring 2002 — Assignment #1 Solutions 13 for this region. (Note that the meaning of 1" differs between this problem and the previous
problem: here it is intended to indicate the direction perpendicularly away from the cylin
ders’ axis) (0) For 1' > R2, Qenc1=(+Q — Q)h/L = 0, giving E = 0. (d) The electron will experience an inwardlydirected electrostatic force with magnitude
given by
w 2 7T60L(R1 + R2) This force will provide the centripetal acceleration for the maintenance of the circular orbit,
so that erE mes2 (&+&W‘
Equating these two expressions for F and recalling that kinetic energy is given by (KE) 2
mpg/2, we obtain
@
(KE)e _ _ 47T60L Fzmeac : END ...
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This note was uploaded on 10/23/2008 for the course 8 8.02 taught by Professor Hudson during the Spring '07 term at MIT.
 Spring '07
 Hudson

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