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 $4·0\times {10}^{-6}\mathrm{C}{\mathrm{m}}^{-2}$. Find the force of attraction between the particle and the plate.

Answer 1

The electric field due to a conducting thin sheet,
$E=\frac{\sigma }{2{\in }_0}$
The magnitude of attractive force between the particle and the plate,
$$\begin{gathered} F\mathit{=}qE \\ F\mathit{=}\frac{\mathit{q}\mathit{\times }\mathit{\sigma }}{\mathit{2}{\mathit{\in }}_{\mathit{0}}} \\ F=\frac{\left(2.0\times {10}^{-6}\right)\times \left(4.0\times {10}^{-6}\right)}{2\times \left(8.85\times {10}^{-12}\right)} \\ F=0.45\mathrm{N} \end{gathered}$$