Here ,
we will find electric field"(E)" due to a charged disc at a distance of "(x=0.01\\ cm=10^{-4}m)" and then
Electric force "=qE"
Given, "\\sigma =20\\ nC\/m^2=20\\times 10^{-6}C\/m^2;x=10^{-4}m;" "R=1\\ m" ";k=\\frac{1}{4\\pi \\epsilon_o}=9\\times 10^9" and "x<<R"
Electric field"(E)= 2\\pi k\\sigma(1-\\frac{x}{\\sqrt{x^2+R^2}})" and "x<<R" .So,
"E=2\\pi k\\sigma (1-\\frac{x}{R})" "=2\\pi \\times 9\\times 10^9\\times 20\\times 10^{-6}(1-\\frac{10^{-4}}{1})\\approx36\\pi \\times 10^4N\/C"
Electric force on the charge"=qE=40\\times 10^{-6}\\times 36\\pi \\times 10^4N=45.2\\ N"
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