Question

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]

Answer 1

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.