(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.0' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 230236, 5986]*) (*NotebookOutlinePosition[ 258166, 6891]*) (* CellTagsIndexPosition[ 258122, 6887]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell[TextData[StyleBox["Ph196bEM Homework 1 Solutions - 2004", \ "Subtitle"]], "Title"], Cell[CellGroupData[{ Cell["1. Jackson (3rd Edition), chapter 1, problem 1.1.", "Section"], Cell[TextData[{ "Use Green's theorem [and (1,21) if necessary] to prove the following:\na) \ Any excess charge placed on a conductor must be entirely on its surface. (A \ conductor by definition contains charges capable of moving freely under the \ action of applied electric fields.)\nb) A closed hollow conductor shields \ its interior from fields due to charges outside, but does not shield its \ exterior from fields due to charges placed inside it.\nc) The electric field \ at the surface of a conductor is normal to the surface and has a magnitude ", Cell[BoxData[ \(TraditionalForm\`\[Sigma]/\[CurlyEpsilon]\_0\)]], ", where \[Sigma] is the charge density per unit area on the surface." }], "Text"], Cell[CellGroupData[{ Cell["Solution Jackson Problem 1.1", "Subsection"], Cell[TextData[{ "a) If there were charge inside a conductor, then by Gauss' law there would \ have to be an electric field inside it also. But this would produce a \ current, i.e., moving charges, and so the problem would not be \ electrostatics. This argument does not apply to the surface of a conductor, \ where charges can indeed collect. Physically, there must be a finite depth \ within which the idealized classical \"surface charge\" resides; its \ calculation is intrinsically quantum mechanical since the atomic structure of \ the conductor must be taken into account. The depth is of order the size of \ atoms, i.e. Angstroms.\nb) For a closed hollow conductor that has no charges \ in its hollow, the potential inside the hollow must be a constant by the \ uniqueness theorem of electrostatics. Note that we have a case in which a \ volume is defined by an equipotential and so one solution is that the whole \ of the volume is at that potential. Since this is A solution, it is THE \ solution by the uniqueness theorem. Thus the volume inside a closed hollow \ conductor is shielded from the electric field of any charge outside the \ conductor. If there is charge inside the volume, there must be an electric \ field in the hollow by Gauss' law. Further, application of Gauss' law to any \ surface completely inside the metal and enclosing the hollow yields zero \ since ", Cell[BoxData[ \(TraditionalForm\`E = 0\)]], " inside the conductor. Thus any charge in the hollow induces an equal and \ opposite charge on the inside surface. The field inside the hollow terminates \ on this inside surface, creating a surface charge density on it.\nc) There \ can be no tangential component of electric field on the surface of a \ conductor since, by Stoke's Theorem applied to ", Cell[BoxData[ \(TraditionalForm\`\(\(\[Del]\&\[RightVector]\(\(\[Cross]\)\(E\&\ \[RightVector]\)\) = 0\)\(,\)\(\ \)\)\)]], " that would imply a tangential field just inside the surface and a \ concomitant tangential current. This does not constrain the normal component \ of the field, however. Apply ", Cell[BoxData[ \(TraditionalForm\`\[Del]\&\[RightVector]\(\(\[CenterDot]\)\(E\&\ \[RightVector]\)\) = \(\(\[Rho]/\[CurlyEpsilon]\_0\)\(\ \)\)\)]], " to a small volume with negligible thickness normal to the surface and \ relatively large flats with area ", Cell[BoxData[ \(TraditionalForm\`\(\[DifferentialD]A\)\&\[RightVector]\)]], " in the direction of the outward normal to the surface on the outside and \ , ", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[\(\(\ \)\(-\(\[DifferentialD]A\)\&\[RightVector]\)\), "TraditionalForm"], " "}], TraditionalForm]]], " on the inside. Then, use the fact that there is no field inside the \ conductor to get ", Cell[BoxData[ \(TraditionalForm\`\[Sigma] = \(\[CurlyEpsilon]\_0\) E\_n\)]], " as the surface charge density where ", Cell[BoxData[ \(TraditionalForm\`E\_n\)]], " is the component of ", Cell[BoxData[ \(TraditionalForm\`\(\(E\&\[RightVector]\)\(\ \)\)\)]], "in the direction of the outward normal to the surface. \n\ ______________________________________________________________________________\ _________" }], "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["2. Lines of Force", "Section"], Cell[TextData[{ "An electric field is often represented by a map of its lines of force. \nA \ line of force is defined by a curve ", Cell[BoxData[ \(TraditionalForm\`\(r\&\[RightVector]\)(s)\)]], " satisfying, at each point where ", Cell[BoxData[ \(TraditionalForm\`E = \(\(|\)\(E\&\[RightVector]\)\(|\)\)\)]], " is finite and non-zero, ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\(\[DifferentialD]\ r\_i\/\[DifferentialD]\ s\)\(=\)\), "TraditionalForm"], \(E\_i\/E\)}], TraditionalForm]]], ". A point were ", Cell[BoxData[ \(TraditionalForm\`E = 0\)]], " is called a neutral point and there is no unique line of force through \ it; similarly at a point charge ", Cell[BoxData[ \(TraditionalForm\`q\)]], " there is not a unique line of force. You can think of lines of force \ beginning or ending at these special points. \nFurther a tube of flux in a \ charge-free region is defined by an open surface ", Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalS]\)]], " with perimeter \[ScriptCapitalP] containing no neutral points or point \ charges and the set of all lines of force through the points of \ \[ScriptCapitalP]. The flux in the tube is defined by ", Cell[BoxData[ FormBox[ RowBox[{"\[ScriptCapitalF]", "=", RowBox[{\(\[Integral]\_\[ScriptCapitalS]\), " ", RowBox[{\(E\&\[RightVector]\[CenterDot]\[DifferentialD]A\&\ \[RightVector]\), " ", "with", " ", Cell[TextData[{ "\[DifferentialD]", Cell[BoxData[ \(TraditionalForm\`A\&\[RightVector]\)]] }]]}]}]}], TraditionalForm]]], "defined over ", Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalS]\)]], " with one sense or the other; it is easy to see that the flux through any \ cross section ", Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalS]\^\[Prime]\)]], " of the tube is just \[ScriptCapitalF] as no flux passes into or out of \ the tube through the wall defined by the lines of force. \nIn geometrically \ symmetric situations, tubes of flux are often easy to calculate and may be \ used to simplify the calculation of lines of force.\n As a simple example, a \ cone of half-angle \[Theta] with vertex on a point charge ", Cell[BoxData[ \(TraditionalForm\`q\)]], " carries a flux of ", Cell[BoxData[ \(TraditionalForm\`q\/\(2\ \[CurlyEpsilon]\_0\)\)]], Cell[BoxData[ \(TraditionalForm\`\((1 - cos\ \[Theta])\)\)]], " and the lines of force are radial from the charge. You can think of the \ surface \[ScriptCapitalS] as a spherical cap over the end of a truncated cone \ centered at the origin having a radius \[Epsilon] of arbitrarily small value.\ \n\nSuppose ", Cell[BoxData[ \(TraditionalForm\`q\_0 = 1\)]], " is located at the origin and ", Cell[BoxData[ \(TraditionalForm\`q\_1 = 2\)]], " is located at ", Cell[BoxData[ \(TraditionalForm\`r\_1 = \(\((0, 0, \(-1\))\)\(.\)\)\)]], " A line of force makes an angle \[Theta] with the +", Cell[BoxData[ \(TraditionalForm\`z\)]], "-axis as it approaches the origin. Find the angle \[Psi] that this line \ makes with the +", Cell[BoxData[ \(TraditionalForm\`z\)]], "-axis as it recedes to infinity. \n" }], "Text"], Cell[CellGroupData[{ Cell["Solution - Lines of Force", "Subsection"], Cell[TextData[{ "A tube of flux starting on a point charge ", Cell[BoxData[ \(TraditionalForm\`q\_o = 1\)]], " with angle \[Theta] to the +", Cell[BoxData[ \(TraditionalForm\`z\)]], "-axis at the origin, as indicated in the (carefully calculated) sketch, \ carries a flux of ", Cell[BoxData[ \(TraditionalForm\`\(1\/\(2\ \[CurlyEpsilon]\_0\)\) \((1 - cos\ \[Theta])\)\)]], ". (Strictly speaking, the tube of flux is defined by the flux through a \ spherical cap of infinitesimal radius from the origin capping the truncated \ cone of half angle \[Theta].) " }], "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Lines of Force for ", Cell[BoxData[ \(TraditionalForm\`q\_0 = 1\)]], " at origin and ", Cell[BoxData[ \(TraditionalForm\`q\_1 = 2\)]], " at {0,0,-1}. 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Cell[TextData[{ "By simple superposition, the flux through a disk whose normal is the ", Cell[BoxData[ \(TraditionalForm\`z\)]], "-axis and whose radius is \[Rho] is just\n\n", Cell[BoxData[ \(TraditionalForm\`\(1\/\(2\ \[CurlyEpsilon]\_0\)\) \((1 - z\/\@\(z\^2 + \[Rho]\^2\))\)\)]], "+", Cell[BoxData[ \(TraditionalForm\`\(2\/\(2\ \[CurlyEpsilon]\_0\)\) \((1 - \(z + \ 1\)\/\@\(\((z + 1)\)\^2 + \[Rho]\^2\))\)\)]], "\n\nand the point ", Cell[BoxData[ \(TraditionalForm\`\(\((z, \[Rho])\)\(\ \)\)\)]], "will be on the tube of flux if this is equal to ", Cell[BoxData[ \(TraditionalForm\`\(1\/\(2\ \[CurlyEpsilon]\_0\)\) \((1 - cos\ \[Theta])\)\)]], ". As the line of force approaches infinity we have both " }], "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`z\/\@\(z\^2 + \[Rho]\^2\) \[Rule] cos\ \[Psi]\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\(z + 1\)\/\@\(\((z + 1)\)\^2 + \[Rho]\^2\) \[Rule] \ \ cos\ \(\(\[Psi]\)\(.\)\)\)]], " " }], "Text"], Cell["Therefore", "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\((1 - cos\ \[Theta])\) = \(3 \((1 - cos\ \[Psi])\) = \(2\ \(sin\^2\) \[Theta]\/2 = 6\)\)\)]], Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\(\(sin\^2\) \[Psi]\/2\)\(,\)\(\ \ \)\)\)\)]] }], "Text"], Cell["or", "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\[Psi] = 2\ \(\(\(sin\^\(-1\)\)(\(1\/\@3\) \(sin\^2\) \ \[Theta]\/2)\)\(.\)\)\)]], "\nAlthough it was not asked for, it is easy to see from the \"sketch\" \ (and to show analytically) that the lines of force point back to the neutral \ point on the ", Cell[BoxData[ \(TraditionalForm\`z\)]], " axis at ", Cell[BoxData[ \(TraditionalForm\`z = \(-\@2\) + 1. \)]] }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["3. Jackson, Ch. 1 problem 1.4.", "Section"], Cell[TextData[{ "Each of three charged spheres of radius ", Cell[BoxData[ \(TraditionalForm\`a\)]], ", one conducting, one having a uniform charge density within its volume, \ and one having a spherically symmetric charge density that varies radially as \ ", Cell[BoxData[ \(TraditionalForm\`r\^n\)]], " (", Cell[BoxData[ \(TraditionalForm\`n > \(-3\)\)]], "), has a total charge ", Cell[BoxData[ \(TraditionalForm\`Q\)]], ". Use Gauss' s theorem to obtain the electric fields both inside and \ outside each sphere. Sketch the behavior of the fields as a function of \ radius for the first two spheres, and for the third with ", Cell[BoxData[ \(TraditionalForm\`n = \(-1\), \ \(+2. \)\)]] }], "Text"], Cell[CellGroupData[{ Cell["Solution - Jackson problem 1.4.", "Subsection"], Cell[TextData[{ "Since any rotation about the origin in any of the three cases given leaves \ the physical situation unchanged, the electric field must have only a radial \ component. For any fixed radius ", Cell[BoxData[ \(TraditionalForm\`r \[NotEqual] a\)]], ", we have from Maxwell's equation,\n", Cell[BoxData[ \(TraditionalForm\`\[Del]\&\[RightVector]\(\(\[CenterDot]\)\(E\&\ \[RightVector]\)\) = \[Rho]\/\[CurlyEpsilon]\_\(\(\ \)\(0\)\)\)]], ",\nGauss's law\n", Cell[BoxData[ \(TraditionalForm\`\[Integral]\_V\[Del]\&\[RightVector]\(\(\[CenterDot]\ \)\(E\&\[RightVector]\)\) \[DifferentialD]\^3 x\ = \ \[Integral]\_\(\[PartialD]V\)E\&\[RightVector]\ \[CenterDot]\(\(\[DifferentialD]A\&\[RightVector]\)\(\ \)\)\)]], ",\nand symmetry that \n", Cell[BoxData[ \(TraditionalForm\`E( r) = \(Q(r)\)\/\(4 \[Pi]\ \(\[CurlyEpsilon]\_\(\(\ \)\(0\)\)\) \ r\^2\)\)]], ",\nwhere ", Cell[BoxData[ \(TraditionalForm\`\(\(E(r)\)\(\ \)\)\)]], "is the radial component of ", Cell[BoxData[ \(TraditionalForm\`E\&\[RightVector]\)]], " at radius ", Cell[BoxData[ \(TraditionalForm\`r\)]], " and ", Cell[BoxData[ \(TraditionalForm\`Q(r)\)]], " is the amount of charge inside the radius ", Cell[BoxData[ \(TraditionalForm\`r\)]], ". This relation may also be used at surface ", Cell[BoxData[ \(TraditionalForm\`r = a\)]], " if the electric field is continuous there. It is not continuous in the \ conducting sphere case there is a delta function of charge on the surface), \ but it is in the other two (no delta function of charge on the ", Cell[BoxData[ \(TraditionalForm\`r = a\)]], " surface). " }], "Text"], Cell[CellGroupData[{ Cell["Conducting sphere case", "Subsubsection"], Cell[TextData[{ "From the above, we get ", Cell[BoxData[ \(TraditionalForm\`E(r) = 0\)]], " for ", Cell[BoxData[ \(TraditionalForm\`r < a\)]], " since there is no charge inside the sphere (all charge is on its outside \ surface) and\n ", Cell[BoxData[ FormBox[ RowBox[{\(E(r)\), "=", RowBox[{\(Q\/\(4 \[Pi]\ \(\[CurlyEpsilon]\_\(\(\ \)\(0\)\)\) r\^2\)\), Cell[""]}]}], TraditionalForm]]], "for ", Cell[BoxData[ \(TraditionalForm\`r > a\)]], ". \n \n It is usual to define the electric field at ", Cell[BoxData[ \(TraditionalForm\`r = a\)]], " as the limit\n\n", Cell[BoxData[ FormBox[ RowBox[{\(E(a)\), "=", RowBox[{\(lim\+\(r \[Rule] 0, \ r > 0\)E(r)\), "=", RowBox[{Cell[""], Cell[""]}]}]}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`Q\/\(4 \[Pi]\ \(\[CurlyEpsilon]\_\(\(\ \)\(0\)\)\) \ a\^2\)\)]], "\n" }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Uniform Charge Distribution", "Subsubsection"], Cell[TextData[{ "The charge density \[Rho] is zero for ", Cell[BoxData[ \(TraditionalForm\`r > a\)]], " and\n\n ", Cell[BoxData[ \(TraditionalForm\`\(3\ Q\)\/\(4 \[Pi]\ a\^3\)\)]], "\n \n for ", Cell[BoxData[ \(TraditionalForm\`r < a\)]], ". Thus ", Cell[BoxData[ \(TraditionalForm\`Q( r) = \(\(\(3\ Q\)\/\(4 \[Pi]\ a\^3\)\) \(\(\(\ \)\(4\ \[Pi]\)\)\/3\ \) r\^3 = Q\ \((r\/a)\)\^3\)\)]], " for ", Cell[BoxData[ \(TraditionalForm\`r < a\)]], ". Thus the radial electric field is\n \n ", Cell[BoxData[ \(TraditionalForm\`E( r) = \(Q\/\(4 \[Pi]\ \(\[CurlyEpsilon]\_\(\(\ \)\(0\)\)\) r\^2\)\ \ \((r\/a)\)\^3\ = \ \(\(\(Q\)\(\ \)\)\/\(4 \[Pi]\ \(\[CurlyEpsilon]\_\(\(\ \ \)\(0\)\)\) a\^2\)\) r\/a\)\)]], " for ", Cell[BoxData[ \(TraditionalForm\`r < a\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\(\(Q\)\(\ \)\)\/\(4 \[Pi]\ \(\[CurlyEpsilon]\_\(\(\ \ \)\(0\)\)\) r\^2\)\)]], " for ", Cell[BoxData[ \(TraditionalForm\`r > \(\(a\)\(.\)\)\)]] }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Power Law Charge Density", "Subsubsection"], Cell[TextData[{ "If ", Cell[BoxData[ \(TraditionalForm\`\[Rho] = C\ \((r\/a)\)\^n\)]], " for ", Cell[BoxData[ \(TraditionalForm\`n > \(-3\)\)]], ", and the total charge inside radius ", Cell[BoxData[ \(TraditionalForm\`a\)]], " is ", Cell[BoxData[ \(TraditionalForm\`Q\)]], ", then\n\n", Cell[BoxData[ \(TraditionalForm\`Q = \(\[Integral]\_0\%a C\ \(\((r\/a)\)\^n\) 4\ \[Pi]\ \(r\^2\) \[DifferentialD]r = \(4\ \[Pi]\ C\ a\^\(-n\)\ \ \(\[Integral]\_0\%a\ \(r\^\(\(\ \)\(n + 2\)\)\) \[DifferentialD]r\) = \(\(4\ \[Pi]\ C\)\/\(n \ + 3\)\) a\^\(-n\)\ \(\(a\^\(n + 3\)\)\(\ \)\(.\)\)\)\)\)]], "\n\nNote that the logarithmic possibility if ", Cell[BoxData[ \(TraditionalForm\`n\)]], " were ", Cell[BoxData[ \(TraditionalForm\`\(-3\)\)]], " is avoided by the ", Cell[BoxData[ \(TraditionalForm\`n > \(-3\)\)]], " condition. Solve this for ", Cell[BoxData[ \(TraditionalForm\`C\)]], " to get" }], "Text"], Cell[TextData[{ " ", Cell[BoxData[ \(TraditionalForm\`\[Rho] = \(\((n + 3)\) Q\)\/\(4\ \[Pi]\ a\^3\)\ \((r\ \/a)\)\^n\)]], "." }], "Text"], Cell["So", "Text"], Cell[TextData[Cell[BoxData[ \(TraditionalForm\`Q( r) = \(\[Integral]\_0\%r\(\((n + 3)\) Q\)\/\(4\ \[Pi]\ a\^3\)\ \(\((r\ \/a)\)\^n\) 4\ \[Pi]\ \(r\^2\) \[DifferentialD]r = Q\ \((r\/a)\)\^\(n + 3\)\)\)]]], "Text"], Cell["and ", "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`E( r) = \(Q\/\(4 \[Pi]\ \(\[CurlyEpsilon]\_\(\(\ \)\(0\)\)\) r\^2\)\ \ \((r\/a)\)\^\(n + 3\)\ = \ \(\(\(Q\)\(\ \)\)\/\(4 \[Pi]\ \ \(\[CurlyEpsilon]\_\(\(\ \)\(0\)\)\) a\^2\)\) \((r\/a)\)\^\(n + 1\)\ for\ r < a\ and\ \(\(Q\)\(\ \)\)\/\(4 \[Pi]\ \(\[CurlyEpsilon]\_\(\(\ \ \)\(0\)\)\) r\^2\)\ for\ r > a\)\)]], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Plot[{UnitStep[r - 1]\/r\^2, If[r < 1, r, 1\/r\^2], If[r < 1, r\^\(-1\), 1\/r\^2], If[r < 1, r\^3, 1\/r\^2]}, 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Jackson, Ch. 1 problem 1.12.", "Section"], Cell[TextData[{ "Prove ", StyleBox["Green's reciprocation theorem: ", FontSlant->"Italic"], "If \[CapitalPhi] is the potential due to a volume-charge density", StyleBox[" ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`\[Rho]\)]], " within a volume ", Cell[BoxData[ \(TraditionalForm\`V\)]], " and the surface-charge density \[Sigma] on the conducting surface ", Cell[BoxData[ \(TraditionalForm\`S\)]], " bounding the volume ", Cell[BoxData[ \(TraditionalForm\`V\)]], ", while ", Cell[BoxData[ \(TraditionalForm\`\[CapitalPhi]\^\[Prime]\)]], " is the potential due to another charge distribution ", Cell[BoxData[ \(TraditionalForm\`\[Rho]\^\[Prime]\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[Sigma]\^\[Prime]\)]], ", then" }], "Text"], Cell[TextData[{ " ", Cell[BoxData[ \(TraditionalForm\`\[Integral]\_\(\(V\)\(\ \)\)\[Rho]\ \ \(\[CapitalPhi]\^\[Prime]\) \[DifferentialD]\^3 r + \[Integral]\_\(\(S\)\(\ \)\)\[Sigma]\ \(\[CapitalPhi]\^\ \[Prime]\) \[DifferentialD]A = \[Integral]\_\(\(V\)\(\ \)\)\[Rho]\^\[Prime]\ \ \[CapitalPhi] \[DifferentialD]\^3 r + \[Integral]\_\(\(S\)\(\ \)\)\[Sigma]\^\[Prime]\ \ \[CapitalPhi] \[DifferentialD]A\)]] }], "Text"], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution - Green's Reciprocation Theorem, Jackson \ Problem 1.12", FontWeight->"Bold"]], "Subsection"], Cell[TextData[{ "Let \[CapitalPhi] be the potential in a volume ", Cell[BoxData[ \(TraditionalForm\`V\)]], " produced by a space charge density ", Cell[BoxData[ \(TraditionalForm\`\[Rho]\)]], " in ", Cell[BoxData[ \(TraditionalForm\`V\)]], " and a surface charge density ", Cell[BoxData[ \(TraditionalForm\`\[Sigma]\)]], " on the ", StyleBox["conducting", FontSlant->"Italic"], " surface ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]V\)]], "of ", Cell[BoxData[ \(TraditionalForm\`V\)]], ". Then we know that inside ", Cell[BoxData[ \(TraditionalForm\`V\)]] }], "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\[Del]\^2 \[CapitalPhi] = \ \(-\[Rho]\)/\[CurlyEpsilon]\_0\)]], "." }], "Text"], Cell[TextData[{ "In exactly the same geometry, let ", Cell[BoxData[ \(TraditionalForm\`\[CapitalPhi]\^\[Prime]\)]], " be produced by ", Cell[BoxData[ \(TraditionalForm\`\[Rho]\^\[Prime]\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[Sigma]\^\[Prime]\)]], " so that inside ", Cell[BoxData[ \(TraditionalForm\`V\)]] }], "Text"], Cell[TextData[Cell[BoxData[ \(TraditionalForm\`\[Del]\^2 \[CapitalPhi]\^\[Prime] = \ \(-\[Rho]\^\[Prime]\)/\(\(\[CurlyEpsilon]\_0\)\(.\)\)\)]]], "Text"], Cell["\<\ Then, \"cross multiplying\" and subtracting these equations yields\ \>", "Text"], Cell[TextData[Cell[BoxData[ \(TraditionalForm\`\[CapitalPhi]\^\[Prime]\ \[Del]\^2 \[CapitalPhi]\ - \ \ \[CapitalPhi]\ \[Del]\^2 \[CapitalPhi]\^\[Prime]\ = \(-\ \ \(1\/\[CurlyEpsilon]\_0\)\) \(\((\[CapitalPhi]\^\[Prime]\ \[Rho]\ \ - \ \ \[CapitalPhi]\ \[Rho]\^\[Prime]\ )\)\(.\)\(\ \ \)\)\)]]], "Text"], Cell["But, ", "Text"], Cell[TextData[Cell[BoxData[ \(TraditionalForm\`\[Del]\&\[RightVector]\(\(\[CenterDot]\)\(\ \)\((\ \[CapitalPhi]\^\[Prime]\ \[Del]\&\[RightVector] \[CapitalPhi])\)\) = \[Del]\&\ \[RightVector] \[CapitalPhi]\^\[Prime]\[CenterDot]\ \[Del]\&\[RightVector] \ \[CapitalPhi]\ + \ \[CapitalPhi]\^\[Prime]\ \[Del]\^2 \[CapitalPhi]\)]]], \ "Text"], Cell["and similarly ", "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\[Del]\&\[RightVector]\(\(\[CenterDot]\)\(\ \)\((\ \[CapitalPhi]\ \[Del]\&\[RightVector] \[CapitalPhi]\^\[Prime])\)\) = \[Del]\&\ \[RightVector] \[CapitalPhi]\[CenterDot]\ \[Del]\&\[RightVector] \ \[CapitalPhi]\^\[Prime]\ + \ \[CapitalPhi]\ \[Del]\^2 \ \[CapitalPhi]\^\[Prime]\)]], "." }], "Text"], Cell["Subtracting, the symmetric term disappears and so we get", "Text"], Cell[TextData[Cell[BoxData[ \(TraditionalForm\`\[CapitalPhi]\^\[Prime]\ \[Del]\^2 \[CapitalPhi]\ - \ \ \[CapitalPhi]\ \[Del]\^2 \[CapitalPhi]\^\[Prime] = \ \(\[Del]\&\[RightVector]\ \(\(\[CenterDot]\)\(\ \)\((\[CapitalPhi]\^\[Prime]\ \[Del]\&\[RightVector] \ \[CapitalPhi] - \[CapitalPhi]\ \[Del]\&\[RightVector] \ \[CapitalPhi]\^\[Prime])\)\) = \(-\(1\/\[CurlyEpsilon]\_0\)\) \(\((\ \[CapitalPhi]\^\[Prime]\ \[Rho]\ \ - \ \[CapitalPhi]\ \[Rho]\^\[Prime]\ \ )\)\(.\)\)\)\)]]], "Text"], Cell[TextData[{ "Next integrate the divergence term and the last term over just the inside \ of ", Cell[BoxData[ \(TraditionalForm\`V\)]], " to get" }], "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\[Integral]\_\(\[PartialD]V\)\ \((\[CapitalPhi]\^\ \[Prime]\ \[Del]\&\[RightVector] \[CapitalPhi] - \[CapitalPhi]\ \[Del]\&\ \[RightVector] \[CapitalPhi]\^\[Prime])\)\[CenterDot]\[DifferentialD]A\&\ \[RightVector]\ = \(-\(1\/\[CurlyEpsilon]\_0\)\)\ \(\[Integral]\_V\((\ \[CapitalPhi]\^\[Prime]\ \[Rho]\ \ - \ \[CapitalPhi]\ \[Rho]\^\[Prime]\ )\) \ \[DifferentialD]\^3 x\)\)]], " (eq 1)." }], "Text"], Cell["\<\ Then, use the fact that the surface charge density on the conductor satisfies \ \ \>", "Text"], Cell[TextData[Cell[BoxData[ \(TraditionalForm\`\[Sigma]\/\[CurlyEpsilon]\_0 = \(n\&\[RightVector]\ \[CenterDot] E\&\[RightVector] = \(-n\&\[RightVector]\)\[CenterDot]\[Del]\&\ \[RightVector] \[CapitalPhi]\)\)]]], "Text"], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`n\&\[RightVector]\)]], " is a unit vector pointing ", StyleBox["away", FontSlant->"Italic"], " from the conductor and so ", StyleBox["into", FontSlant->"Italic"], " the volume ", Cell[BoxData[ \(TraditionalForm\`V\)]], ". Thus for the case at hand, ", Cell[BoxData[ \(TraditionalForm\`\(-n\&\[RightVector]\)\)]], " is a unit vector in the direction of the ", Cell[BoxData[ \(TraditionalForm\`\(\(\[DifferentialD]A\&\[RightVector]\)\(\ \)\)\)]], "in eq. 1. We finally get," }], "Text"], Cell[TextData[Cell[BoxData[ \(TraditionalForm\`\(1\/\[CurlyEpsilon]\_0\) \ \(\[Integral]\_\(\[PartialD]V\)\ \((\[CapitalPhi]\^\[Prime]\ \[Sigma] - \ \[CapitalPhi]\ \[Sigma]\^\[Prime])\) \[DifferentialD]A\)\ = \(-\(1\/\ \[CurlyEpsilon]\_0\)\)\ \(\[Integral]\_V\((\[CapitalPhi]\^\[Prime]\ \[Rho]\ \ \ - \ \[CapitalPhi]\ \[Rho]\^\[Prime]\ )\) \[DifferentialD]\^3 x\)\)]]], "Text"], Cell["or", "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\[Integral]\_V \[CapitalPhi]\^\[Prime]\ \ \[Rho]\ \ \[DifferentialD]\^3 x\ + \ \[Integral]\_\(\[PartialD]V\)\ \ \[CapitalPhi]\^\[Prime]\ \[Sigma] \[DifferentialD]A\ = \ \[Integral]\_V \ \[CapitalPhi]\ \(\(\[Rho]\)\(\ \)\)\^\[Prime]\ \[DifferentialD]\^3 x\ + \ \[Integral]\_\(\[PartialD]V\)\ \[CapitalPhi]\ \(\ \[Sigma]\^\[Prime]\) \(\(\[DifferentialD]A\)\(\ \)\(.\)\)\)\)\)]], " " }], "Text"], Cell[TextData[{ "Notice that the surface ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]V\)]], " may consist of several different conductors, each at a different \ potential. This and similar results have interesting and useful application \ to different electrical situations in a particular geometry. " }], "Text"], Cell[TextData[{ "An example is that the geometry is a pair of coaxial cylinders od radii ", Cell[BoxData[ \(TraditionalForm\`a\)]], " and ", Cell[BoxData[ \(TraditionalForm\`b\)]], ", ", Cell[BoxData[ \(TraditionalForm\`b > a\)]], ". In the unprimed case, the outer cylinder is grounded, the inner is at \ potential ", Cell[BoxData[ \(TraditionalForm\`V\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`\[Rho] = 0. \)]], " In the primed case, the potentials are the same, but the charge density \ is a unit point charge at radius ", Cell[BoxData[ \(TraditionalForm\`c\)]], ", ", Cell[BoxData[ \(TraditionalForm\`a < c < \(\(b\)\(.\)\(\ \)\)\)]], "This is the situation in a cylindrical ionization chamber." }], "Text"], Cell[TextData[{ " ", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(V \(\[Integral]\_\(inner\ cylinder\)\ \ \ \[Sigma] \[DifferentialD]A\)\ = \ \[CapitalPhi]\ \((c)\) + \ V \(\[Integral]\_\(inner\ cylinder\)\ \(\[Sigma]\^\[Prime]\) \ \[DifferentialD]A\)\)\)\)]] }], "Text"], Cell["\<\ Thus, we can get the amount of charge induced on the inner cylinder by the \ point charge between the plates:\ \>", "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\[Integral]\_\(inner\ cylinder\)\ \ \[Sigma] \ \[DifferentialD]A\ - \ \[Integral]\_\(inner\ cylinder\)\ \ \(\[Sigma]\^\[Prime]\) \[DifferentialD]A\ = \ \(\(charge\ induced\ on\ inner\ \ cylinder\)\(\ \)\(=\)\(\ \)\)\)]], Cell[BoxData[ \(TraditionalForm\`\(\[CapitalPhi]\ \((c)\)\)\/V\)]] }], "Text"], Cell["\<\ and calculating the right hand side is very easy. Thus as the charge moves \ between the plates, you can easily determine, as a function of time, the \ current that flows in an external circuit maintaining the potential.\ \>", "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["5.Jackson, Ch. 2 problem 2.1, subsectons a, b, c, d, f.", "Section"], Cell["A point charge \[Theta]", "Text"], Cell[TextData[{ "Let the infinite grounded conductor be in the ", Cell[BoxData[ \(TraditionalForm\`\(\(z\)\(=\)\(0\)\(\ \)\)\)]], "plane and let the charge ", Cell[BoxData[ \(TraditionalForm\`q\)]], " be on the ", Cell[BoxData[ \(TraditionalForm\`\(\(z\)\(-\)\(axis\)\(\ \)\)\)]], "at ", Cell[BoxData[ \(TraditionalForm\`z = d > 0\)]], ". Then we can calculate the field in the ", Cell[BoxData[ \(TraditionalForm\`z \[GreaterEqual] 0\)]], " region as the sum of the field due to ", Cell[BoxData[ \(TraditionalForm\`q\)]], " itself and the field of a charge ", Cell[BoxData[ \(TraditionalForm\`\(-q\)\)]], " located on the ", Cell[BoxData[ \(TraditionalForm\`z - axis\)]], " at ", Cell[BoxData[ \(TraditionalForm\`z = \(-d\)\)]], ". It is very important to notice that the field due to the \"image\" \ charge is not due to an actual charge at the indicated image position, but is \ the cumulative effect of the distributed induced charge on the conducting \ plate. This charge has moved in from infinity (in our idealization) as the \ exciting charge ", Cell[BoxData[ \(TraditionalForm\`q\)]], " was moved into place." }], "Text"], Cell[TextData[{ "The potential at the point ", Cell[BoxData[ \(TraditionalForm\`\((z \[GreaterEqual] 0, \ \[Rho], \[CurlyPhi])\) = \((x = \(\[Rho]\ cos\ \[CurlyPhi] . \ \ y = \[Rho]\ sin\ \[CurlyPhi]\), \ z \[GreaterEqual] 0)\)\)]], " is" }], "Text"], Cell[BoxData[ \(\[CapitalPhi][ z_, \[Rho]_] := \(1\/\(4 \[Pi]\ \[CurlyEpsilon]\_0\)\) \((q\/\@\(\ \[Rho]\^2 + \((z - d)\)\^2\) - q\/\@\(\[Rho]\^2 + \((z + d)\)\^2\))\)\)], "Input", CellLabel->"In[8]:="], Cell[TextData[{ "It satisfies the boundary conditions ", Cell[BoxData[ \(TraditionalForm\`\[CapitalPhi](z = 0, \[Rho], \[CurlyPhi]) = 0, \ \[CapitalPhi] = 0\ at\ \[Infinity], \)]], " and in the region ", Cell[BoxData[ \(TraditionalForm\`z \[GreaterEqual] 0\)]], " it satisfies Poisson's equation for a point charge at ", Cell[BoxData[ \(TraditionalForm\`\((\[Rho] = 0, \ z = d)\)\)]], " and so by the uniqueness theorem, it is the potential in the region ", Cell[BoxData[ \(TraditionalForm\`z \[GreaterEqual] 0\)]], ".\n\na) The surface charge density ", Cell[BoxData[ \(TraditionalForm\`\(\(\[Sigma](\[Rho])\)\(\ \)\)\)]], "on the plate satisfies ", Cell[BoxData[ \(TraditionalForm\`\(\[Sigma](\[Rho])\)\/\[CurlyEpsilon]\_0 = \(-\ \[PartialD]\_z\ \(\[CapitalPhi](z, \[Rho])\)\)\( | \_\(z = 0\)\) . \)]] }], "Text"], Cell[BoxData[ \(\[Sigma][\[Rho]_] := \(-\[CurlyEpsilon]\_0\) \((\[PartialD]\_z\ \ \[CapitalPhi][z, \[Rho]])\) /. z \[Rule] 0\)], "Input", CellLabel->"In[9]:="], Cell[CellGroupData[{ Cell[BoxData[ \(\[Sigma][\[Rho]]\)], "Input", CellLabel->"In[10]:="], Cell[BoxData[ \(\(-\(\(d\ q\)\/\(2\ \[Pi]\ \((d\^2 + \[Rho]\^2)\)\^\(3/2\)\)\)\)\)], \ "Output", CellLabel->"Out[10]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Plot[ Evaluate[{\[Sigma][\[Rho]] /. {d \[Rule] .5}, \[IndentingNewLine]\ \[Sigma][\[Rho]] /. {d \[Rule] 1}, 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]], Cell["Of course, it is just the value of the image charge.", "Text"], Cell[TextData[{ "b) The force on the charge ", Cell[BoxData[ \(TraditionalForm\`q\)]], " is in the ", Cell[BoxData[ \(TraditionalForm\`\(-z\)\)]], " direction and its magnitude is " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Fq = \(1\/\(4 \[Pi]\ \[CurlyEpsilon]\_0\)\) q\^2\/\((2 d)\)\^2\)], "Input", CellLabel->"In[13]:="], Cell[BoxData[ \(q\^2\/\(16\ d\^2\ \[Pi]\ \[CurlyEpsilon]\_0\)\)], "Output", CellLabel->"Out[13]="] }, Open ]], Cell[TextData[{ "Alternatively, calculate the force (in the ", Cell[BoxData[ \(TraditionalForm\`\(+z\)\)]], " direction) by integrating over the contributions of all of the induced \ charge on the plate" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Fq2 = \(\(2 \[Pi]\ q\)\/\(4 \[Pi]\ \[CurlyEpsilon]\_0\)\) \(\ \[Integral]\_0\%\[Infinity] \[Sigma][\[Rho]] \(d\/\((d\^2 + \ \[Rho]\^2)\)\^\(3/2\)\) \[Rho] \[DifferentialD]\[Rho]\)\)], "Input", CellLabel->"In[14]:="], Cell[BoxData[ \(\(-\(q\^2\/\(16\ d\^2\ \[Pi]\ \[CurlyEpsilon]\_0\)\)\)\)], "Output", CellLabel->"Out[14]="] }, Open ]], Cell["Of course, this yields the same result.", "Text"], Cell[TextData[{ "c) Alternatively, we can calculate the force acting on the plane using the \ fact that the force on an element of surface charge ", Cell[BoxData[ \(TraditionalForm\`\[Sigma]\ \[DifferentialD]A\)]], " is ", Cell[BoxData[ \(TraditionalForm\`\(\[Sigma]\^2\/\(2\ \[CurlyEpsilon]\_0\)\) \ \[DifferentialD]A\)]], " acting normal to the surface and away from it. Thus the force on the \ plate is in the ", Cell[BoxData[ \(TraditionalForm\`\(+z\)\)]], " direction and has magnitude" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Fplate = \(\(2 \[Pi]\)\/\(2 \[CurlyEpsilon]\_0\)\) \(\[Integral]\_0\%\ \[Infinity]\( \[Sigma][\[Rho]]\^2\) \[Rho] \[DifferentialD]\[Rho]\)\)], \ "Input", CellLabel->"In[15]:="], Cell[BoxData[ \(q\^2\/\(16\ d\^2\ \[Pi]\ \[CurlyEpsilon]\_0\)\)], "Output", CellLabel->"Out[15]="] }, Open ]], Cell["\<\ which is of course the expected value, being equal and opposite to the force \ on the charge. The factor of 2 in the expression for the outward stress on a charged \ conductor seems a bit mysterious since one is tempted to write the force \ without it using Coulomb's law. However, the electric field here is due to \ the charge density itself, and a little more care is necessary. Here is how \ to get the result.\ \>", "Text"], Cell[TextData[{ "Think of the surface charge as being actually distributed slightly into \ the conductor and not a mathematical delta function on its surface. Then \ letting ", Cell[BoxData[ \(TraditionalForm\`z\)]], " be distance into the conductor, ", Cell[BoxData[ \(TraditionalForm\`z = 0\)]], " at the surface, we can write from ", Cell[BoxData[ \(TraditionalForm\`\[Del]\&\[RightVector]\(\(\[CenterDot]\)\(E\&\ \[RightVector]\)\) = \[Rho]\/\[CurlyEpsilon]\_0\)]], "that ", Cell[BoxData[ \(TraditionalForm\`\(\[CurlyEpsilon]\_0\) \[DifferentialD]E\_z\/\ \[DifferentialD]\ z = \[Rho](z)\)]], ". The force on the conductor due to the charge in ", Cell[BoxData[ \(TraditionalForm\`\[DifferentialD]z\ \[DifferentialD]A\)]], " is ", Cell[BoxData[ FormBox[ RowBox[{\(\[DifferentialD]F\_z\), "=", RowBox[{\(\[DifferentialD]A\ \(\(E\_z\)(z)\)\ \(\[Rho]( z)\)\ \[DifferentialD]z\), "=", RowBox[{\(\[DifferentialD]A\ \(\[CurlyEpsilon]\_0\) \(\(E\_z\)( z)\) \(\[DifferentialD]E\_z\/\[DifferentialD]\ z\) \[DifferentialD]z\), "=", RowBox[{\(\[DifferentialD]A\), \(1\/2\), \(\[CurlyEpsilon]\_0\), " ", \(\[DifferentialD]\((\(\(E\_z\)(z)\)\^2)\)\), RowBox[{Cell[""], "."}]}]}]}]}], TraditionalForm]]], " Thus the total force per unit area is ", Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\[Infinity]\( 1\/2\) \(\[CurlyEpsilon]\_0\) \ \[DifferentialD]\((\(\(E\_z\)(z)\)\^2)\)\ = \ \(\(1\/2\) \[CurlyEpsilon]\_0\ \ E\_surface\%2\ = 1\/\(2\ \[CurlyEpsilon]\_0\)\ \(\(\[Sigma]\^2\)\(.\)\)\)\)]], " " }], "Text"], Cell[TextData[{ "d) The work necessary to move the charge from ", Cell[BoxData[ \(TraditionalForm\`z\)]], " to ", Cell[BoxData[ \(TraditionalForm\`z + \[DifferentialD]z\)]], " is ", Cell[BoxData[ \(TraditionalForm\`\[DifferentialD]W = \(\(q\^2\) \[DifferentialD]z\)\/\ \(16\ z\^2\ \[Pi]\ \[CurlyEpsilon]\_0\)\)]], ". The work to move it from ", Cell[BoxData[ \(TraditionalForm\`z = d\)]], " to \[Infinity] is" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_d\%\[Infinity]\( q\^2\/\(16\ z\^2\ \[Pi]\ \ \[CurlyEpsilon]\_0\)\) \[DifferentialD]z\)], "Input", CellLabel->"In[16]:="], Cell[BoxData[ \(q\^2\/\(16\ d\ \[Pi]\ \[CurlyEpsilon]\_0\)\)], "Output", CellLabel->"Out[16]="] }, Open ]], Cell[TextData[{ "f) To get the work in electron volts, we set ", Cell[BoxData[ \(TraditionalForm\`\(\(q\)\(\ \)\)\)]], "to be the electron charge, ", Cell[BoxData[ \(TraditionalForm\`e\)]], ", then divide out one factor of ", Cell[BoxData[ \(TraditionalForm\`e\)]], " to become the \"", Cell[BoxData[ \(TraditionalForm\`electron\)]], "\" in electron-Volt, and then evaluate in SI units producing Volts.\n" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(e\/\(16\ d\ \[Pi]\ \[CurlyEpsilon]\_0\) /. {e \[Rule] 1.6\ 10\^\(-19\), d \[Rule] 10\^\(-10\), \[CurlyEpsilon]\_0 \[Rule] 1\/\(9\ 10\^16\ 4 \[Pi]\ 10\^\(-7\)\)}\)], "Input", CellLabel->"In[17]:="], Cell[BoxData[ \(3.6`\)], "Output", CellLabel->"Out[17]="] }, Open ]], Cell["So the result is 3.6 eV.", "Text", CellLabel->"In[18]:="] }, Open ]], Cell[CellGroupData[{ Cell["6. Jackson, Ch. 2 problem 2.2.", "Section"], Cell[TextData[{ "For a point charge at radius ", Cell[BoxData[ \(TraditionalForm\`h\)]], " on the ", Cell[BoxData[ \(TraditionalForm\`\(\(z\)\(-\)\(axis\)\(\ \)\)\)]], "inside a grounded sphere of radius ", Cell[BoxData[ \(TraditionalForm\`a\)]], ", ", Cell[BoxData[ \(TraditionalForm\`0 < h < a\)]], ", we consider an image charge ", Cell[BoxData[ \(TraditionalForm\`\(-q\^\[Prime]\)\)]], " at radius ", Cell[BoxData[ \(TraditionalForm\`h\^\[Prime] > a\)]], " on the ", Cell[BoxData[ \(TraditionalForm\`z - axis\)]], ". Then we adjust the two parameters ", Cell[BoxData[ \(TraditionalForm\`q\^\[Prime]\)]], " and ", Cell[BoxData[ \(TraditionalForm\`h\^\[Prime]\)]], "so that the sphere at radius ", Cell[BoxData[ \(TraditionalForm\`a\)]], " is at zero potential (if possible - it is not ", StyleBox["a priori", FontSlant->"Italic"], " obvious that this will work!)" }], "Text"], Cell[CellGroupData[{ Cell["Start graphics definitions", "Subsubsection"], Cell[BoxData[ \(<< Graphics`Arrow`\)], "Input", CellLabel->"In[18]:="], Cell[CellGroupData[{ Cell[BoxData[ \(sphericalimage = Graphics[{Circle[{0, 0}, 1], Line[{{0, 0}, {0, 2}}], PointSize[ .02], Point[{0, 0}], Point[{0, .6}], Point[{0, 1/ .6}], Point[{ .5, \@\(1 - .25\)}], Arrow[{0, 0}, { .5, \@ .75}], Arrow[{0, .6}, { .5, \@ .75}], Arrow[{0, 1/ .6}, { .5, \@ .75}], Arrow[{0, 0}, {0, .6}], Arrow[{0, 0}, {0, 1/ .6}], Text[\*"\"\<\!\(h\&\[RightVector]\)\>\"", {\(- .07\), .3}, {1, 0}], Text[\*"\"\<\!\(h\&\[RightVector]\^\[Prime]\)\>\"", {\(- \ .07\), 1.2}, {1, 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ool005l0oooo00<000000?ooo`3oool0GP3oool005l0oooo00<000000?ooo`3oool0GP3oool005l0 oooo00<000000?ooo`3oool0GP3oool005l0oooo00<000000?ooo`3oool0GP3oool005l0oooo00<0 00000?ooo`3oool0GP3oool005l0oooo00<000000?ooo`3oool0GP3oool005l0oooo00<000000?oo o`3oool0GP3oool00<00oooo00300?ooo`00`03oool00<00oooo00300?ooo`00`03oool00<00oooo 0000\ \>"], ImageRangeCache->{{{125.688, 316.688}, {613.75, 326.75}} -> {-2.43193, \ 2.51481, 0.0109949, 0.0109949}}], Cell[BoxData[ TagBox[\(\[SkeletonIndicator] Graphics \[SkeletonIndicator]\), False, Editable->False]], "Output", CellLabel->"Out[20]="] }, Open ]], Cell[TextData[{ "Notice that ", Cell[BoxData[ \(TraditionalForm\`\(\(|\)\(r\&\[RightVector]\)\(|\)\)\ = \ \(\(a\)\(.\ \)\)\)]] }], "Text"], Cell[TextData[{ "The potential at point ", Cell[BoxData[ \(TraditionalForm\`r\&\[RightVector]\)]], " is " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Phi] = \(1\/\(4 \[Pi]\ \[CurlyEpsilon]\_0\)\) \((q\/d - q\^\[Prime]\/d\^\[Prime])\)\)], "Input", CellLabel->"In[21]:="], Cell[BoxData[ FractionBox[ RowBox[{\(q\/d\), "-", FractionBox[ SuperscriptBox["q", "\[Prime]", MultilineFunction->None], SuperscriptBox["d", "\[Prime]", MultilineFunction-> None]]}], \(4\ \[Pi]\ \[CurlyEpsilon]\_0\)]], "Output", CellLabel->"Out[21]="] }, Open ]], Cell[TextData[{ "But ", Cell[BoxData[ \(TraditionalForm\`d = \(\(\(|\)\(r\&\[RightVector] - h\&\[RightVector]\)\(|\)\) = \(\(|\)\(a\ e\&\[RightVector]\_r - h e\&\[RightVector]\_z\)\(|\)\)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`d\^\[Prime] = \(\(\(|\)\(r\&\[RightVector] - d\&\[RightVector]\^\[Prime]\)\(|\)\) = \(\(|\)\(a\ e\&\ \[RightVector]\_r - \(h\^\[Prime]\) e\&\[RightVector]\_z\)\(|\)\)\)\)]], ". So we have" }], "Text"], Cell[BoxData[ \(\[Phi] = \(1\/\(4 \[Pi]\ \[CurlyEpsilon]\_0\)\) \((1\/\(\(|\)\(a\/q\ e\ \&\[RightVector]\_r - \(h\/q\) e\&\[RightVector]\_z\)\(|\)\) - 1\/\(\(|\)\(a\/q\^\[Prime]\ e\&\[RightVector]\_r - \ \(h\^\[Prime]\/q\^\[Prime]\) e\&\[RightVector]\_z\)\(|\)\))\)\)], "Text"], Cell[TextData[{ "Now use the fact that ", Cell[BoxData[ \(TraditionalForm\`\(\(|\)\(a\ u\&\[RightVector] + b\ v\&\[RightVector]\)\(|\)\) = \(\(\(|\)\(b\ u\&\[RightVector] + a\ v\&\[RightVector]\)\(|\)\) = \@\(a\^2 + b\^2 + 2 a\ b\ u\&\ \[RightVector]\[CenterDot]v\&\[RightVector]\)\)\)]], " if ", Cell[BoxData[ \(TraditionalForm\`u\&\[RightVector]\)]], " and ", Cell[BoxData[ \(TraditionalForm\`v\&\[RightVector]\)]], " are unit vectors. So we can write " }], "Text"], Cell[BoxData[ \(\[Phi] = \(1\/\(4 \[Pi]\ \[CurlyEpsilon]\_0\)\) \((1\/\(\(|\)\(a\/q\ e\ \&\[RightVector]\_r - \(h\/q\) e\&\[RightVector]\_z\)\(|\)\) - 1\/\(\(|\)\(h\^\[Prime]\/q\^\[Prime]\ e\&\[RightVector]\_r - \ \(a\/q\^\[Prime]\) e\&\[RightVector]\_z\)\(|\)\))\)\)], "Text"], Cell[TextData[{ "and this will vanish if ", Cell[BoxData[ \(TraditionalForm\`a/q = h\^\[Prime]/q\^\[Prime]\)]], " and ", Cell[BoxData[ \(TraditionalForm\`h/q = \(\(a/q\^\[Prime]\)\(\ \)\)\)]], "or ", Cell[BoxData[ \(TraditionalForm\`q\^\[Prime] = \(a\/h\) q\)]], " and ", Cell[BoxData[ \(TraditionalForm\`h\^\[Prime] = \(a\ q\^\[Prime]\/q = a\^2\/h\)\)]], "." }], "Text"], Cell[TextData[{ "a) The potential at a point ", Cell[BoxData[ \(TraditionalForm\`\(\(r\&\[RightVector]\)\(=\)\)\)]], " inside the sphere is" }], "Text"], Cell[BoxData[ \(\[CapitalPhi][ r_, \[Theta]_] := \(q\/\(4 \[Pi]\ \[CurlyEpsilon]\_0\)\) \((1\/\@\(r\ \^2 + h\^2 - 2 r\ h\ Cos[\[Theta]]\) - \(a\/h\) 1\/\@\(r\^2 + a\^4/h\^2 - 2 r\ \(a\^2\/h\) \ Cos[\[Theta]]\))\)\)], "Input", CellLabel->"In[22]:="], Cell[TextData[{ "Check for typing errors by checking the boundary condition that the sphere \ of radius ", Cell[BoxData[ \(TraditionalForm\`a\)]], " is at zero potential." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[\[CapitalPhi][a, \[Theta]], {a > 0, h > 0}]\)], "Input", CellLabel->"In[23]:="], Cell[BoxData[ \(0\)], "Output", CellLabel->"Out[23]="] }, Open ]], Cell["\<\ b) The induced surface charge on the inside surface of the grounded sphere is\ \ \>", "Text"], Cell[BoxData[ \(\[Sigma][\[Theta]_] := \[CurlyEpsilon]\_0\ \((\[PartialD]\_r\ \ \[CapitalPhi][r, \[Theta]])\) /. r \[Rule] a\)], "Input", CellLabel->"In[24]:="], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[\[Sigma][\[Theta]], {a > 0, h > 0}]\)], "Input", CellLabel->"In[25]:="], Cell[BoxData[ \(\(\((\(-a\^2\) + h\^2)\)\ q\)\/\(4\ a\ \[Pi]\ \((a\^2 + h\^2 - 2\ a\ h\ \ Cos[\[Theta]])\)\^\(3/2\)\)\)], "Output", CellLabel->"Out[25]="] }, Open ]], Cell[TextData[{ "Check the trivial case that ", Cell[BoxData[ \(TraditionalForm\`h \[Rule] 0\)]], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[% /. h \[Rule] 0, {a > 0, h > 0}]\)], "Input", CellLabel->"In[26]:="], Cell[BoxData[ \(\(-\(q\/\(4\ a\^2\ \[Pi]\)\)\)\)], "Output", CellLabel->"Out[26]="] }, Open ]], Cell[TextData[{ "More generally, check the integral of the surface charge density, which \ should be ", Cell[BoxData[ \(TraditionalForm\`\(-q\)\)]], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[\[Integral]\_0\%\[Pi] \[Sigma][\[Theta]]\ 2 \[Pi]\ \(a\^2\) Sin[\[Theta]] \[DifferentialD]\[Theta], {a > h, h > 0}]\)], "Input",\ CellLabel->"In[27]:="], Cell[BoxData[ \(\(-q\)\)], "Output", CellLabel->"Out[27]="] }, Open ]], Cell[TextData[{ "c) Clearly we can get the force on ", Cell[BoxData[ \(TraditionalForm\`q\)]], " from the image charge (since the cumulative effect of the surface charge \ on the inside of the sphere is just the field on the inside of the sphere \ produced by the image), so it is along the ", Cell[BoxData[ \(TraditionalForm\`\(+z\) - axis\)]], " with magnitude\n\n", Cell[BoxData[ \(TraditionalForm\`F = \(\(q\^2\/\(4 \[Pi]\ \[CurlyEpsilon]\_0\)\) \(a\ \/h\) 1\/\((h\^\[Prime] - h)\)\^2 = \(\(q\^2\/\(4 \[Pi]\ \ \[CurlyEpsilon]\_0\)\) \(a\ h\)\/\((a\^2 - h\^2)\)\^2 = \(q\^2\/\(4 \[Pi]\ \ \[CurlyEpsilon]\_0\)\) \(a\ h\)\/\(\(\((a - h)\)\^2\) \((a + \ h)\)\^2\)\)\)\)]], "\n\nwhich goes to zero as ", Cell[BoxData[ \(TraditionalForm\`h \[Rule] 0\)]], "and to ", Cell[BoxData[ \(TraditionalForm\`\(q\^2\/\(4 \[Pi]\ \[CurlyEpsilon]\_0\)\) 1\/\(4 \((a - h)\)\^2\)\)]], "as ", Cell[BoxData[ \(TraditionalForm\`h \[Rule] a\)]], ", which is the plane sheet case. We can also get the result by integrating \ over the charge density to get the force in the ", Cell[BoxData[ \(TraditionalForm\`\(+z\)\)]], " direction. To help ", StyleBox["Mathematica", FontSlant->"Italic"], " do the integral in reasonable time, make the substitution ", Cell[BoxData[ \(TraditionalForm\`Cos[\[Theta]] \[Rule] x\)]], " and do a bit of pre-simplification on the integrand." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[\(-\(\(\(\ \)\(q\)\)\/\(4 \[Pi]\ \[CurlyEpsilon]\_0\)\)\) 2 \[Pi] \(\[Integral]\_\(-1\)\%1 Evaluate[\(Simplify[\[Sigma][\[Theta]] /. Cos[\[Theta]] \[Rule] x, {a > h, h > 0}]\/d\^2\) \(a\ x - h\)\/d /. d -> \@\(a\^2 + h\^2 - 2 a\ h\ x\)] \(a\^2\) \ \[DifferentialD]x\), {a > h, h > 0}]\)], "Input", CellLabel->"In[28]:="], Cell[BoxData[ \(\(a\ h\ q\^2\)\/\(4\ \((a - h)\)\^2\ \((a + h)\)\^2\ \[Pi]\ \ \[CurlyEpsilon]\_0\)\)], "Output", CellLabel->"Out[28]="] }, Open ]], Cell["As above.", "Text"], Cell[TextData[{ "d) If the sphere is set to potential ", Cell[BoxData[ \(TraditionalForm\`V\)]], " then, inside it, the potential is what was calculated in a) above plus ", Cell[BoxData[ \(TraditionalForm\`V\)]], ". Notice that this satisfies the boundary condition that the inside \ surface is an equipotential ", Cell[BoxData[ \(TraditionalForm\`V\)]], " and the Poisson equation is satisfied inside, and, being A solution it is \ THE solution by the uniqueness theorem." }], "Text"], Cell[TextData[{ "Similarly, if a charge ", Cell[BoxData[ \(TraditionalForm\`Q\)]], " is added to the conductor, it becomes an equipotential ", Cell[BoxData[ \(TraditionalForm\`V\^\[Prime]\)]], " and so inside the potential is a) above plus ", Cell[BoxData[ \(TraditionalForm\`\(\(V\^\[Prime]\)\(.\)\)\)]], " The value of ", Cell[BoxData[ \(TraditionalForm\`V\^\[Prime]\)]], "is determined by the value of ", Cell[BoxData[ \(TraditionalForm\`Q\)]], " (all of which is on the outside surface) and the shape of the ", StyleBox["OUTSIDE", FontSlant->"Italic"], " surface of the conductor. 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Cell[CellGroupData[{ Cell[StyleData["Note"], CellFrame->True, CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, CellFrameColor->RGBColor[0, 0, 1], ShowSpecialCharacters->Automatic, Hyphenation->True, LineSpacing->{1, 3}, ParagraphSpacing->{0, 8}, CounterIncrements->"Text", FontFamily->"Helvetica", FontSize->10], Cell[StyleData["Note", "Printout"], CellMargins->{{7, 4}, {0, 8}}, CellFrameColor->GrayLevel[0], FontSize->8] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Picture"], CellMargins->{{10, Inherited}, {0, 8}}, CellHorizontalScrolling->True], Cell[StyleData["Picture", "Printout"], CellMargins->{{7, Inherited}, {0, 8}}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Caption"], CellMargins->{{10, 50}, {0, 3}}, PageBreakAbove->False, Hyphenation->True, FontFamily->"Helvetica", FontSize->9, FontColor->RGBColor[0, 0, 1]], Cell[StyleData["Caption", "Printout"], CellMargins->{{7, 50}, {2, 4}}, FontSize->7, FontColor->GrayLevel[0]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Tables", "Subsection"], Cell[CellGroupData[{ Cell[StyleData["2ColumnTable"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, GridBoxOptions->{ColumnWidths->{0.39, 0.59}, ColumnAlignments->{Left}}], Cell[StyleData["2ColumnTable", "Printout"], CellMargins->{{7, 4}, {0, 8}}, FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["3ColumnTable"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, StyleMenuListing->None, GridBoxOptions->{ColumnWidths->0.325, ColumnAlignments->{Left}}], Cell[StyleData["3ColumnTable", "Printout"], CellMargins->{{7, 4}, {0, 8}}, FontSize->10] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Formulas and Programming", "Subsection"], Cell[CellGroupData[{ Cell[StyleData["ChemicalFormula"], CellMargins->{{55, Inherited}, {Inherited, Inherited}}, CellHorizontalScrolling->True, DefaultFormatType->DefaultInputFormatType, "TwoByteSyntaxCharacterAutoReplacement"->True, 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Cell[StyleData["Program"], CellMargins->{{10, Inherited}, {Inherited, Inherited}}, CellHorizontalScrolling->True, Hyphenation->False, LanguageCategory->"Formula", FontFamily->"Courier"], Cell[StyleData["Program", "Printout"], CellMargins->{{7, Inherited}, {Inherited, Inherited}}, FontSize->9.5] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Hyperlink Styles", "Subsection"], Cell["\<\ The cells below define styles useful for making hypertext ButtonBoxes. The \ \"Hyperlink\" style is for links within the same Notebook, or between \ Notebooks.\ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["Hyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->GrayLevel[1], Background->RGBColor[0, 0.6, 1], ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2]}]&), Active->True, ButtonFrame->"None", ButtonNote->ButtonData}], Cell[StyleData["Hyperlink", "Printout"], FontColor->GrayLevel[0], Background->GrayLevel[1]] }, Closed]], Cell["\<\ The following styles are for linking automatically to the on-line help \ system.\ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["MainBookLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->GrayLevel[1], Background->RGBColor[0, 0.6, 1], ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "MainBook", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["MainBookLink", "Printout"], FontColor->GrayLevel[0], Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["AddOnsLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Courier", FontColor->GrayLevel[1], Background->RGBColor[0, 0.6, 1], ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "AddOns", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["AddOnLink", "Printout"], FontColor->GrayLevel[0], Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["RefGuideLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Courier", FontColor->GrayLevel[1], Background->RGBColor[0, 0.6, 1], ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "RefGuideLink", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["RefGuideLink", "Printout"], FontColor->GrayLevel[0], Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["GettingStartedLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->GrayLevel[1], Background->RGBColor[0, 0.6, 1], ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "GettingStarted", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["GettingStartedLink", "Printout"], FontColor->GrayLevel[0], Background->GrayLevel[1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["OtherInformationLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->GrayLevel[1], Background->RGBColor[0, 0.6, 1], ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "OtherInformation", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["OtherInformationLink", "Printout"], FontColor->GrayLevel[0], Background->GrayLevel[1]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Palette Styles", "Subsection"], Cell["\<\ The cells below define styles that define standard ButtonFunctions, for use \ in palette buttons.\ \>", "Text"], Cell[StyleData["Paste"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, After]}]&)}], Cell[StyleData["Evaluate"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], SelectionEvaluate[ FrontEnd`InputNotebook[ ], All]}]&)}], Cell[StyleData["EvaluateCell"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], FrontEnd`SelectionMove[ FrontEnd`InputNotebook[ ], All, Cell, 1], FrontEnd`SelectionEvaluateCreateCell[ FrontEnd`InputNotebook[ ], All]}]&)}], Cell[StyleData["CopyEvaluate"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionCreateCell[ FrontEnd`InputNotebook[ ], All], FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], FrontEnd`SelectionEvaluate[ FrontEnd`InputNotebook[ ], All]}]&)}], Cell[StyleData["CopyEvaluateCell"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionCreateCell[ FrontEnd`InputNotebook[ ], All], FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], FrontEnd`SelectionEvaluateCreateCell[ FrontEnd`InputNotebook[ ], All]}]&)}] }, Closed]] }, Open ]] }] ] (******************************************************************* Cached data follows. 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