Show that the expression for the electric field of a charged disk showing in the equation below reduces to that of a point for Z>>R, where all symbols have their usual meanings.
σ/2εo[ 1= z/√z^2+R^2]
The electric field of a charged disk
"E=\\frac{\\sigma}{2\\epsilon_0}(1-\\frac{z}{\\sqrt{z^2+R^2}})=\\frac{\\sigma}{2\\epsilon_0}(1-\\frac{1}{\\sqrt{1+\\frac{R^2}{z^2}}})"
For "z\\gg R"
"\\frac{1}{\\sqrt{1+\\frac{R^2}{z^2}}}\\approx1-\\frac{R^2}{2z^2}" . So, we have
"E=\\frac{\\sigma}{2\\epsilon_0}(1-\\frac{1}{\\sqrt{1+\\frac{R^2}{z^2}}})\\approx\\frac{\\sigma R^2}{4\\epsilon_0h^2}=\\frac{q}{4\\pi\\epsilon_0z^2}" as for the point charge field.
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