Follow the Charge

$The\quad problem\quad essentially\quad talks\quad about\quad equipotential\quad surfaces.\\ If\quad the\quad charge\quad was\quad placed\quad at\quad an\quad arbitrary\quad point\quad in\quad space\quad and\quad then\quad moved\quad around\quad in\quad such\quad a\quad way\quad that\quad requires\quad no\quad energy\quad then\quad \\ that\quad path\quad must\quad lie\quad on\quad an\quad equipotential\quad surface\quad because\quad the\quad electric\quad field\quad is\quad perpendicular\quad to\quad all\quad points\quad on\quad such\quad a\quad surface\quad and\quad \\ thus\quad the\quad work\quad to\quad be\quad done\quad against\quad it\quad would\quad be\quad zero.\\ Thus\quad we\quad have\quad to\quad find\quad the\quad equation\quad of\quad the\quad equipotential\quad surface\quad that\quad has\quad potential\quad V_{ 0 }.\\ The\quad general\quad formula\quad for\quad potential\quad at\quad any\quad point\quad can\quad be\quad calculated\quad conventionally\quad by\quad calculating\quad the\quad individual\quad potentials\\ due\quad to\quad each\quad charge,\quad summing\quad them,\quad and\quad then\quad solving\quad the\quad problem\quad according\quad to\quad the\quad conditions\quad given.\quad \\ However,\quad this\quad is\quad very\quad tedious\quad and\quad can\quad be\quad avoided\quad if\quad we\quad make\quad a\quad very\quad interesting\quad observation.\\ The\quad Octahedral\quad arrangement\quad of\quad charges\quad essentially\quad consists\quad of\quad three\quad pairs\quad of\quad \\ opposite\quad charges\quad each\quad separated\quad by\quad a\quad distance\quad of\quad 2\Delta \quad (By\quad Geometry\quad of\quad the\quad Octahedron)\quad and\quad all\quad pairs\quad are\quad perpendicular\quad to\quad each\quad other\quad \\ and\quad meet\quad at\quad a\quad common\quad point.\\ If\quad we\quad assume\quad this\quad common\quad point\quad to\quad be\quad the\quad origin\quad of\quad a\quad co-ordinate\quad system,\quad we\quad can\quad then\quad assign\quad the\quad three\quad axes\quad along\quad the\quad three\quad pairs\quad of\quad charges.\\ Furthermore,\quad since\quad the\quad distance\quad \Delta <<1\quad and\quad the\quad initial\quad point\quad is\quad at\quad a\quad sufficiently\quad large\quad distance\quad from\quad the\quad Octahedron,\quad the\quad pairs\quad of\quad charges\\ can\quad be\quad approximated\quad to\quad behave\quad as\quad dipoles.\\ Thus,\quad we\quad end\quad up\quad with\quad three\quad dipoles,\quad each\quad oriented\quad along\quad a\quad particular\quad axis.\\ Now,\quad we\quad already\quad know\quad the\quad potential\quad of\quad a\quad dipole\quad oriented\quad along\quad the\quad z-axis,\quad V=\frac { Q(d)\cos { { \theta }_{ z } } }{ 4\pi \Omega { R }^{ 2 } } \\ where,\quad \cos { { \theta }_{ z } } =\frac { z }{ R } \quad and\quad { R }^{ 2 }=\left( x^{ 2 }+y^{ 2 }+z^{ 2 } \right) =\Gamma \\ Similarly,\quad calculating\quad the\quad potentials\quad due\quad to\quad the\quad other\quad two\quad dipoles\quad and\quad taking\quad d=2\Delta ,\\ The\quad total\quad potential\quad \Phi =\frac { (Q\Delta )(x+y+z) }{ 2\pi \Omega { \Gamma }^{ 1.5 } } \\ We\quad already\quad know\quad that\quad we\quad want\quad the\quad equation\quad for\quad the\quad surface\quad that\quad is\quad equipotential\quad for\quad potential\quad { V }_{ 0 }.\\ Thus,\quad we\quad set\quad \Phi ={ V }_{ 0 }\\ This\quad will\quad give\quad us\quad the\quad required\quad implicit\quad equation\quad for\quad the\quad surface\quad we\quad want\\ \because \quad \Phi ={ V }_{ 0 }\\ \therefore \quad \frac { (Q\Delta )(x+y+z) }{ 2\pi \Omega { \Gamma }^{ 1.5 } } ={ V }_{ 0 }\\ Which\quad on\quad proper\quad rearrangement\quad gives,\\ \boxed { \quad z\quad =\quad \Gamma \left[ \frac { { (2\pi { V }_{ 0 }\Omega ) }^{ 2 }{ \Gamma }^{ 2 }-{ (Q\Delta ) }^{ 2 } }{ 2{ (Q\Delta ) }^{ 2 }(x+y) } \right] -\frac { xy }{ (x+y) } \quad } \\$