A charged partical having a charge of `-2.0 xx 10^(-6) C` is placed close to a nonconducting plate having a surface charge density `4.0 xx 10^(6) C m^(-2). Fine the force of attraction between the particle and the plate.

Explanation

We know that E due to conducting charge
density
`E = sigma/2 epsilon_0`
Then, ` F =qE = ((q sigma)/ (2 epsilon_0))`
` = ((2.0 xx (10^-6) xx 4.0 xx (10^-6))/(2 xx 8.8 xx (10^-12))`
` =45N.`