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
Answer & explanation
Correct answer: option 1
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)