PSet 1 Answers

PSet 1 Answers - Adam S. Bolton February 15, 2002...

Info iconThis preview shows pages 1–13. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13
This is the end of the preview. Sign up to access the rest of the document.

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 x-axis, 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: — 0.4|3x (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 " . |x|3 |x—0.4|3 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 (+3/372-7/(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 21-49.) 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 y-component, 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 21-9): 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 A-2. 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 fE-dA : (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 21-65.) -Q +Q —l/2 l/2 7" Let the charges be located on the x-axis, 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 x-axis 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 E-field is proportional to 1 /r3 whereas the E-field 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 z-axis 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). | |G|f280 :(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 22-21.) 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 fE-dA : 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 < 1-2 is only possible if 01 : 0, regardless of the value of 02. Problem 1.8 Two concentric charged cylinders. (Giancoli 22-29.) 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 inwardly-directed 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 ...
View Full Document

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.

Page1 / 13

PSet 1 Answers - Adam S. Bolton February 15, 2002...

This preview shows document pages 1 - 13. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online