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

Answer

`0.45 N`

Explanation

Here, `q = -2xx10^(-6) C, sigma = 4xx10^(-6) Cm^(-2)`
Field of attraction between the charged particle and the plate,
`F = qE = (sigma q)/(2 in_(0)) = (4xx10^(-6) xx2xx10^(-6))/(2xx8.845xx10^(-12))`
`= 0.45 N`