Question

Find the intensity of an electric field at a distance of 2× 10^-8 cm from a monovalent ion.The ion is point - charged.

Answer 1

Solution

Distance d =$2*10^{-8}m$

Monovalent charge q=charge of electrons

So

the intensity of an electric field at this distance

$E=\frac{Kq}{d^2}\\=\frac{9*10^9*1.6*10^{-19}}{(2*10^{-8})^2}$

$=3.6*10^6N/C.$