Question

a plant charge q=40nc is placed in front of a circuit disc of 1m radius and at a distance of 0.01cm. the disk carries a positive charge uniformly distributed with a surface charge density of 20nc/m2. what is the force on the charge points?

Answer 1

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$=qE=40\times 10^{-6}\times 36\pi \times 10^4N=45.2\ N$