Question

A charged particle with a charge of −2⋅0 × 10⁻⁶ C is placed close to a non-conducting plate with a surface charge density of 4.0 × 10^-6Cm0^-2. Find the force of attraction between the particle and the plate.

Answer 1

The electric field due to a conducting thin sheet,

`"E" = sigma/( 2 ∈ _0)`

The magnitude of attractive force between the particle and the plate,

F =qE

`"F" = (q xx sigma)/(2∈_0)`

`"F" = ((2.0 xx 10^-6 ) xx ( 4.0 xx 10^-6))/(2 xx (8.55 xx 10^-12))`

F = 0.45 N