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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 3: Electric fields and Capacitors"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.1, Page 79"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "Q = 50e-03;                 # Electric charge, C\n",
      "A = 600e-06;                # Area of plate, m^2\n",
      "\n",
      "#Calculations\n",
      "# Solving for electric field density, D\n",
      "D = Q/A;                      # Electric field density, C/m^2\n",
      "\n",
      "#Result\n",
      "print \"The density of the electric field existing between the plates = %4.1f C/m-square\"%D\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The density of the electric field existing between the plates = 83.3 C/m-square\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.2, Page 80"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "A = 400e-06;              # Cross-sectional area of plate, m^2\n",
      "I = 50e-06;              # Source current, A\n",
      "t = 3;                   # Flow time of current, s\n",
      "\n",
      "#Calculations\n",
      "# Since electric current is the rate of flow of charge i.e I = Q/t, solving for Q \n",
      "Q = I*t;                 # Amount of charge on plates, C\n",
      "#Solving for density of the electric field between the plates\n",
      "D = Q/A;                    # Electric field density, C/m^2\n",
      "\n",
      "#Results\n",
      "print \"The charge on the plates = %3d micro-coloumb\"%(Q/1e-06);\n",
      "print \"The density of the electric field between the plates = %5.3f C/m-square\"%D\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The charge on the plates = 150 micro-coloumb\n",
        "The density of the electric field between the plates = 0.375 C/m-square\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.3, Page 83"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "d = 3e-03;                  # Thickness of dielectric, m\n",
      "Q = 35e-03;                 # Electric charge on plates, C\n",
      "V = 150;                   # Supply voltage, V\n",
      "A = 144e-06;              # Cross-sectional area of plates, m^2\n",
      "\n",
      "#Calculations\n",
      "# Part (a)\n",
      "# Since electric field strength(E) = potential gradient therefore we have\n",
      "E = V/d;                    # Electric field strength, V/m\n",
      "# Part (b)\n",
      "# Solving for electric field density, D\n",
      "D = Q/A;                      # Electric field density, C/m^2\n",
      "\n",
      "#Results\n",
      "print \"The electric field strength = %2d kV/m\"%(E*1e-03);\n",
      "print \"The flux density = %5.1f C/m^2\"%D\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The electric field strength = 50 kV/m\n",
        "The flux density = 243.1 C/m^2\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.4, Page 83"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "d = 4e-03;                   # Thickness of air, m\n",
      "Q = 2e-04;                  # Electric charge on plates, C\n",
      "V = 125;                   # Supply voltage, V\n",
      "D = 15;                   # Electric field density, coulomb-per-metre-square\n",
      "\n",
      "#Calculations\n",
      "# Part (a)\n",
      "# Since electric field strength(E) = potential gradient, therefore we have \n",
      "E = V/d;                    # Electric field strength, V/m\n",
      "# Part (b)\n",
      "# Since D = Q/A, solving for A\n",
      "A = Q/D;              # Cross-sectional area of plates, m^2\n",
      "# Part (c)\n",
      "# Since Q = C*V, solving for C\n",
      "C = Q/V;                    # Capacitance of the plates, F\n",
      "\n",
      "#Results\n",
      "print \"The electric field strength between the plates = %5.2f kV/m\"%(E*1e-03);\n",
      "print \"The csa of the field between the plates = %4.1f mm^2\"%(A/1e-06);\n",
      "print \"The capacitance of the plates = %3.1f micro-coulomb\"%(C/1e-06);\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The electric field strength between the plates = 31.25 kV/m\n",
        "The csa of the field between the plates = 13.3 mm^2\n",
        "The capacitance of the plates = 1.6 micro-coulomb\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.5, Page 86"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "A = 6e-04;                          # Cross-sectional area of plates, m^2\n",
      "d = 5e-04;                          # Thickness of mica sheet, m\n",
      "Epsilon_r = 5.8;                    # Relative permittivity, unitless\n",
      "Epsilon_0 = 8.854e-12;              # Permittivity of Free Space\n",
      "V = 200;                            # Potential difference, V\n",
      "\n",
      "#Calculations\n",
      "# Part (a)\n",
      "# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for d & putting Epsilon = Epsilon_0*Epsilon_r \n",
      "C = ( Epsilon_r*Epsilon_0*A )/d;         # Capacitance, F\n",
      "# Part (b)\n",
      "# Since electric field strength(E) = potential gradient, therefore we have \n",
      "E = V/d;                    # Electric field strength, V/m\n",
      "\n",
      "#Results\n",
      "print \"The capacitance of the capacitor = %5.2f pF\"%(C/1e-12);\n",
      "print \"Electric field strength = %3d kV/m\"%(E*1e-03);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The capacitance of the capacitor = 61.62 pF\n",
        "Electric field strength = 400 kV/m\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.6, Page 86"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "C = 0.224e-09;                    #Capacitance, F\n",
      "A = 5625e-06;                     # Cross-sectional area of plates, m^2\n",
      "Epsilon_r = 2.5;                  # Relative permittivity\n",
      "Epsilon_0 = 8.854e-12;            # Permittivity of Free Space\n",
      "\n",
      "#Calculations\n",
      "# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for d & putting Epsilon = Epsilon_0*Epsilon_r\n",
      "d = ( Epsilon_r*Epsilon_0*A )/C;                  # Thickness of waxed paper dielectric, m\n",
      "\n",
      "#Result\n",
      "print \"The thickness of paper required = %3.2f mm\"%(d/1e-03);\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The thickness of paper required = 0.56 mm\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.7, Page 86"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "C = 4.7e-08;                   #Capacitance, F\n",
      "A = 4e-04;                   # Cross-sectional area of plates, m^2\n",
      "d = 1e-04;                  # Thickness of dielectric, m\n",
      "Epsilon_0 = 8.854e-12;            # Permittivity of Free Space\n",
      "\n",
      "#Calculations\n",
      "# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for Epsilon_r & putting Epsilon = Epsilon_0*Epsilon_r\n",
      "Epsilon_r = (C*d)/(Epsilon_0*A);                  # Relative permittivity\n",
      "\n",
      "#Result\n",
      "print \"Relative permittivity = %4d\"%Epsilon_r\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Relative permittivity = 1327\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.8, Page 87"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "V = 180;                            # Potential difference, V\n",
      "d = 3e-03;                         # Thickness of dielectric, m\n",
      "A = 4.2e-04;                       # Cross-sectional area of plates, m^2\n",
      "Epsilon_r = 3.5;                   # Relative permittivity\n",
      "Epsilon_0 = 8.854e-12;             # Permittivity of Free Space\n",
      "\n",
      "#Calculations\n",
      "# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for C & putting Epsilon = Epsilon_0*Epsilon_r \n",
      "C = ( Epsilon_r*Epsilon_0*A )/d;         # Capacitance, F\n",
      "# Since C = Q/V, solving for Q\n",
      "Q = C*V;                                 # Electric charge, C\n",
      "# Using  D = Q/A,\n",
      "D = Q/A;                                # Electric field density, C/m^2\n",
      "\n",
      "#Results\n",
      "print \"The flux thus produced = %3.2f nC.\"%(Q/1e-09)\n",
      "print \"The flux density thus produced. = %3.2f micro-coulomb-per-metre-square\"%(D/1e-06); \n",
      " "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The flux thus produced = 0.78 nC.\n",
        "The flux density thus produced. = 1.86 micro-coulomb-per-metre-square\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.9, Page 89"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "C_1 = 4.7e-06;                   #Capacitance, F\n",
      "C_2 = 3.9e-06;                   #Capacitance, F\n",
      "C_3 = 2.2e-06;                   #Capacitance, F\n",
      "\n",
      "#Calculations\n",
      "# The resulting capacitance of parallerly connected capacitors is the sum of the individual capacitance present \n",
      "#in the circuit\n",
      "C = C_1 + C_2 + C_3;             # Resulting capacitance  of the circuit, F\n",
      "\n",
      "#Result\n",
      "print \"The resulting capacitance of the combination = %4.1f micro-farad\"%(C/1e-06);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The resulting capacitance of the combination = 10.8 micro-farad\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.10, Page 90"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "C_1 = 6e-06;                    #Capacitance, F\n",
      "C_2 = 4e-06;                   #Capacitance, F\n",
      "V = 150;                      # Supply voltage, V\n",
      "\n",
      "#Calculations\n",
      "# Part (a)\n",
      "# The reciprocal of the resulting capacitance of capacitors connected in series is the sum of the reciprocal \n",
      "#of the individual capacitances present in the circuit i.e 1/C =  1/C1 + 1/C2, solving for C\n",
      "C = ( C_1*C_2 )/(C_1 + C_2);               # Resulting capacitance, F\n",
      "# Part (b)\n",
      "Q = V*C;                                  # Electric charge on the capacitors, C\n",
      "# Part (c)\n",
      "V_1 = Q/C_1;                              # P.d across capacitor C_1, V\n",
      "V_2 = Q/C_2;                              # P.d across capacitor C_2, V\n",
      "\n",
      "#Results\n",
      "print \"The total capacitance of the combination = %3.1f micro-farad\"%(C/1e-06);\n",
      "print \"The charge on each capacitor = %3d micro-coulomb\"%(Q/1e-06);\n",
      "print \"The p.d. developed across %1d micro-farad capacitor = %2d V\"%(C_1/1e-06, V_1);\n",
      "print \"The p.d. developed across %1d micro-farad capacitor = %2d V\"%(C_2/1e-06, V_2);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The total capacitance of the combination = 2.4 micro-farad\n",
        "The charge on each capacitor = 360 micro-coulomb\n",
        "The p.d. developed across 6 micro-farad capacitor = 60 V\n",
        "The p.d. developed across 4 micro-farad capacitor = 90 V\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.11, Page 91"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "C_1 = 3e-06;                    #Capacitance, F\n",
      "C_3 = 12e-06;                   #Capacitance, F\n",
      "C_2 = 6e-06;                    #Capacitance, F\n",
      "V = 400;                        # Supply voltage, V\n",
      "\n",
      "#Calculations\n",
      "# The reciprocal of the resulting capacitance of capacitors connected in series is the sum of the reciprocal of the \n",
      "#individual capacitances present in the circuit i.e 1/C =  1/C1 + 1/C2 + 1/C_3, solving for C\n",
      "C = (C_1 * C_2 * C_3)/( C_1*C_2 + C_2*C_3 + C_3*C_1);               # Resulting capacitance, F\n",
      "Q = V*C;                                  # Electric charge on the capacitors, C\n",
      "# Part (c)\n",
      "V_1 = Q/C_1;                              # P.d across capacitor C_1, V\n",
      "V_2 = Q/C_2;                              # P.d across capacitor C_2, V\n",
      "V_3 =Q/C_3;                               # P.d across capacitor C_2, V\n",
      "\n",
      "#Results\n",
      "print \"P.d across capacitor %1d micro-farad = %5.1f V\"%(C_1/1e-06, V_1);\n",
      "print \"P.d across capacitor %1d micro-farad = %5.1f V\"%(C_2/1e-06, V_2);\n",
      "print \"P.d across capacitor %2d micro-farad = %4.1f V\"%(C_3/1e-06, V_3);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "P.d across capacitor 3 micro-farad = 228.6 V\n",
        "P.d across capacitor 6 micro-farad = 114.3 V\n",
        "P.d across capacitor 12 micro-farad = 57.1 V\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.12, Page 92"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "V = 200;                      # Supply voltage, voltage\n",
      "C_AB = 2.;                # Capacitance across branch AB, micro-farad\n",
      "C_BC = 3.;                 # Capacitance across branch BC, micro-farad\n",
      "C_CD = 6.;                 # Capacitance across branch CD, micro-farad\n",
      "C_EF = 8.;                # Capacitance across branch EF, micro-farad\n",
      "C_BD = 4.;                 # Capacitance across branch EF, micro-farad\n",
      "\n",
      "#Calculations\n",
      "# Part (a)\n",
      "# Since 3-micro-farad & 6-micro-farad  capacitors are in series & the reciprocal of the resulting capacitance of \n",
      "#capacitors connected in series is the sum of the reciprocal of the individual capacitances present in the circuit, \n",
      "#therefore i.e 1/C =  1/C1 + 1/C2 \n",
      "C_BCD = ( C_BC*C_CD )/(C_BC+C_CD);         # Resulting capacitance across branch BCD, micro-farad\n",
      "#Since C_BCD & 4-micro-farad  capacitors are in parallel & the resulting capacitance of parallerly connected capacitors \n",
      "#is the sum of the individual capacitance present in the circuit\n",
      "C_BD = C_BCD + C_BD;         # Resulting capacitance across branch BD, micro-farad\n",
      "#  Since 2-micro-farad & C_BD  capacitors are in series & the reciprocal of the resulting capacitance of capacitors \n",
      "#connected in series is the sum of the reciprocal of the individual capacitances present in the circuit, therefore, \n",
      "#we have\n",
      "C_AD = (C_BD*C_AB)/(C_BD+C_AB);         # Resulting capacitance across branch AD, micro-farad\n",
      "#Since C_AD & C_EF  capacitors are in parallel & the resulting capacitance of parallerly connected capacitors is the \n",
      "#sum of the individual capacitance present in the circuit\n",
      "C = C_AD + C_EF;         # Resulting capacitance of the circuit, micro-farad\n",
      "Q = V*C;                                  # Electric  charge drawn from the supply, C\n",
      "\n",
      "# Part (b)\n",
      "Q_EF = V*C_EF;                        # The charge on the 8 micro-farad capacitor, micro-coulomb\n",
      "\n",
      "# Part (c)\n",
      "Q_AD = Q - Q_EF;                        # The charge on the 4 micro-farad capacitor, C\n",
      "Q_BD = Q_AD;    # Charge in series combination of capacitors, micro-farad\n",
      "# Since Q = C*V, solving for V\n",
      "V_BD = Q_BD/C_BD;                    # The p.d. across the 4F capacitor,V\n",
      "\n",
      "# Part(d)\n",
      "Q_BCD = V_BD*C_BCD;                  # Electric charge across branch BCD, C\n",
      "Q_BC = Q_BCD;                        # Electric charge, C\n",
      "V_BC = Q_BC/C_BC;                    # The p.d. across the 3 micro-farad capacitor\n",
      "\n",
      "#Results\n",
      "print \"The charge drawn from the supply = %3.1f mC\"%(Q/1e+03);\n",
      "print \"The charge on the %1d micro-farad capacitor = %3.1f mC\"%(C_EF, Q_EF/1e+03);\n",
      "print \"The p.d. across the %1d micro-farad capacitor= %2d V\"%(C_BD, V_BD);\n",
      "print \"The p.d. across the %1d micro-farad capacitor = %5.2f V\"%(Q_BC, V_BC);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The charge drawn from the supply = 1.9 mC\n",
        "The charge on the 8 micro-farad capacitor = 1.6 mC\n",
        "The p.d. across the 6 micro-farad capacitor= 50 V\n",
        "The p.d. across the 100 micro-farad capacitor = 33.33 V\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.13, Page 96"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "N = 20;                  # Number of plates in a capacitor\n",
      "A = 6400e-06;            # Cross - sectional area of plate, m^2\n",
      "d = 1.5e-03;             # Distance between plates, m\n",
      "epsilon_r = 6.4;         # Relative permittivity for mica\n",
      "epsilon_o = 8.854e-12;     # Relative permittivity for free space\n",
      "\n",
      "#Calculations\n",
      "# Calculating the capacitance of the capacitor\n",
      "C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d;       # Capacitance, F\n",
      "\n",
      "#Result\n",
      "print \" The capacitance of the capacitor = %3.1f nF\"%(C/1e-09);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The capacitance of the capacitor = 4.6 nF\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.14, Page 96"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "N = 9;                      # Number of plates in a capacitor\n",
      "A = 1200e-06;               # Cross - sectional area of plate, m^2\n",
      "C = 3e-10;                  # Capacitance, F\n",
      "epsilon_r = 5;              # Relative permittivity for mica\n",
      "epsilon_o = 8.854e-12;      # Relative permittivity for free space\n",
      "\n",
      "#Calculations\n",
      "# Using the formula of capacitance, C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d and solving for d, we have\n",
      "d = ((epsilon_o)*(epsilon_r)*A*(N-1))/C;        # Distance between plates, m\n",
      "\n",
      "#Result\n",
      "print \"The thickness of mica between parallel plates of a capacitor = %4.2f mm\"%(d/1e-03);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The thickness of mica between parallel plates of a capacitor = 1.42 mm\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.15, Page 97"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "N = 11;                       # Number of plates in a capacitor\n",
      "r = 25e-03;                   # Radius of circular plate, m\n",
      "A = (math.pi*r**2);                # Cross - sectional area of plate, m^2\n",
      "d = 5e-04;                    # Distance between plates, m\n",
      "epsilon_r = 1;                # Relative permittivity for air\n",
      "epsilon_o = 8.854e-12;        # Relative permittivity for free space\n",
      "\n",
      "#Calculations\n",
      "# Calculating the capacitance of the capacitor\n",
      "C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d;       # Capacitance, F\n",
      "\n",
      "#Result\n",
      "print \" The capacitance of the capacitor = %3.2f pF\"%(C/1e-10);\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The capacitance of the capacitor = 3.48 pF\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.16, Page 99"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "C_1 = 3e-06;                                  # Capacitance, F\n",
      "C_2 = 6e-06;                                  # Capacitance, F\n",
      "V_1 = 250;                                   # Voltage across capacitor C_1, V\n",
      "\n",
      "#Calculations\n",
      "# Since each capacitor will take charge according to its capacitance, so we have\n",
      "Q = C_1*V_1;                                 # Charge on first capacitor C_1, C\n",
      "W_1 = 0.5*C_1*(V_1**2);                       # Energy stored, J\n",
      "# When the two capacitors are connected in parallel the 3 micro-farad will share its charge with 6 micro-farad capacitor. Thus the total charge in the system will remain unchanged, but the total capacitance will now be different\n",
      "C = C_1 + C_2;                              # Total capacitance, F\n",
      "# Since Q = C*V, solving for V\n",
      "V = Q/C;                                    # Voltage across capacitor C_2, V\n",
      "W = 0.5*C*(V**2);                            # Total energy stored by the combination, J\n",
      "\n",
      "#Results\n",
      "print \"The charge and energy stored by %1d micro-F capacitor are %3.2f mC and %5.2f mJ respectively \"%(C_1/1e-06, Q/1e-03 , W_1/1e-03);\n",
      "print \"The p.d. between the plates = %5.2f V\"%V\n",
      "print \"The energy stored by the combination of %1d micro-F and %1d micro-F capacitors = %5.2f mJ\"%(C_1/1e-06, C_2/1e-06, W/1e-03);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The charge and energy stored by 3 micro-F capacitor are 0.75 mC and 93.75 mJ respectively \n",
        "The p.d. between the plates = 83.33 V\n",
        "The energy stored by the combination of 3 micro-F and 6 micro-F capacitors = 31.25 mJ\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.17, Page 100"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "V = 200;                       # Supply voltage, V\n",
      "C_1 = 10e-06;                  # Capacitance, farad\n",
      "C_2 = 6.8e-06;                 # Capacitance, farad\n",
      "C_3 = 4.7e-06;                 # Capacitance, farad\n",
      "\n",
      "#Calculations\n",
      "# Part (a)\n",
      "# Since each capacitor will take charge according to its capacitance, so we have\n",
      "Q_1 = V*C_1;                   # Charge stored on  capacitor C_1, C\n",
      "W_1 = 0.5*C_1*(V**2);           # Energy stored on  capacitor C_1, J\n",
      "# Part (b)\n",
      "# Since C_2 and C_3 are in series and hence, their equivalent capacitance is given by their series combination\n",
      "C_4 = (C_2 * C_3)/(C_2 + C_3);          # Equivalent capacitance of C_2 and C_3, F\n",
      "# Since C_1 and C_4 are in parallel and hence, their equivalent capacitance is given by their parallel combination\n",
      "C = C_1 + C_4;               # Total capacitance of circuit, F\n",
      "# Since Q = C*V, solving for V\n",
      "V_1 = Q_1/C;                         # New p.d across C_1, V\n",
      "W = 0.5*C*(V_1**2);                 # Total energy remaining in the circuit, J\n",
      "energy_used = W_1 - W;             # Energy, J\n",
      "\n",
      "#Results\n",
      "print \"The charge and energy stored by %2d micro-F capacitor are %1d mC and %2.1f J respectively \"%(C_1/1e-06, Q_1/1e-03, W_1);\n",
      "print \"The new p.d across %2d micro-F capacitor = %5.1f V\"%(C_1/1e-06,V_1);\n",
      "print \"The amount of energy used in charging %3.1f micro-F and %3.2f micro-F capacitors from %2d micro-F capacitor = %4.3f J\"%(C_2/1e-06, C_3/1e-06, C_1/1e-06, energy_used/1e-03);\n",
      " "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The charge and energy stored by 10 micro-F capacitor are 2 mC and 0.2 J respectively \n",
        "The new p.d across 10 micro-F capacitor = 156.5 V\n",
        "The amount of energy used in charging 6.8 micro-F and 4.70 micro-F capacitors from 10 micro-F capacitor = 43.495 J\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.18, Page 101"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "V = 400;               # Supply voltage, V\n",
      "E = 0.5e06;            # Dielectric strength, V/m\n",
      "\n",
      "#Calculations\n",
      "# Since E = V/d, solving for d\n",
      "d = V/E;               # Thickness of dielectric, m\n",
      "\n",
      "#Result\n",
      "print \"The minimum thickness of dielectric required = %3.1fmm\"%(d/1e-03);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The minimum thickness of dielectric required = 0.8mm\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.19, Page 101"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "C = 270e-12;              # Capacitance, F\n",
      "A = 60e-04;               # Cross-sectional area of plate, m^2\n",
      "E = 350e03;               # Dielectric strength, V/m\n",
      "epsilon_r = 2.1;           # Relative permittivity\n",
      "epsilon_o = 8.854e-12;     # Permittivity of free space\n",
      "\n",
      "#Calculations\n",
      "# Part (a)\n",
      "# Since formula for capacitance, C = ((epsilon_o)*(eplison_r)*A)/d, solving for d\n",
      "d = ((epsilon_o)*(epsilon_r)*A)/C;      # Thickness of dielectric, m\n",
      "# Part (b)\n",
      "# Since E = V/d, solving for V\n",
      "V = E*d;                                  # Maximum possible working voltage, V\n",
      "\n",
      "#Results\n",
      "print \"The thickness of Teflon sheet required = %5.4f mm\"%(d/1e-03);\n",
      "print \"The maximum possible working voltage for the capacitor = %5.1f V\"%V;\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The thickness of Teflon sheet required = 0.4132 mm\n",
        "The maximum possible working voltage for the capacitor = 144.6 V\n"
       ]
      }
     ],
     "prompt_number": 19
    }
   ],
   "metadata": {}
  }
 ]
}