The value of q if it floats in air is

$\frac{mg \varepsilon_0 \varepsilon_r}{\sigma}$ $\frac{\sigma}{m g \varepsilon_0 \varepsilon_r}$ $\frac{2 m g \varepsilon_0 \varepsilon_r}{\sigma}$ none of these

$\frac{mg \varepsilon_0 \varepsilon_r}{\sigma}$

The magnitude of electric field due to the charged conducting plate is E = $\frac{\sigma}{\varepsilon_r}$ As the charged particle is floating in air (neglecting the buoyant force due to air we obtain) mg = qE $\Rightarrow q=\frac{m g}{E}$ $\Rightarrow q=\frac{m g \varepsilon_r}{\sigma}$ ∴ (A)