Answer to Question #171265 in Electricity and Magnetism for Chirag

Question

Find the electric field ⃗ (E)due to a disk above ‘z’ height from its centre which has a 

constant surface charge σ. Given that disk’s radius is ‘a’.

Answer 1

$dE=\frac{xdq}{4\pi \varepsilon_0 (r^2+x^2)^{3/2}},$

$E(x)=\int_0^a\frac{2\pi r\sigma xdr}{4\pi \varepsilon_0 (r^2+x^2)^{3/2}}=\frac{\sigma}{2\varepsilon_0}(1-\frac{x}{\sqrt{a^2+x^2}}),$

$E(z)=\frac{\sigma}{2\varepsilon_0}(1-\frac{z}{\sqrt{a^2+z^2}}).$