A small charged particle of mass m and charge q is suspended by an insulated thread in front of a very large sheet of charge density σ. The angle made by the thread with the vertical in equilibrium is

$\tan ^{-1}\left(\frac{\sigma q}{2 \varepsilon_0 mg}\right)$

In equilibrium, along x-axis,

T sin θ = qE

⇒ T sin θ = q$\frac{σ}{2ε_0}$          . . . (1)

Where T is the tension in the string.

Along y-axis in equilibrium, T cos θ = mg      . . . (2)

From (1) and (2) we obtain,

$\tan \theta=\frac{q \sigma}{2 \varepsilon_0 mg} \Rightarrow \theta=\tan ^{-1}\left(\frac{\sigma q}{2 \varepsilon_0 mg}\right)$

∴  (A)