Question

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

• A. tan⁻¹$\left( \frac{\sigma q}{2\varepsilon_{0}mg} \right)$
• B. tan ⁻¹$\left( \frac{\sigma q}{\varepsilon_{0}mg} \right)$
• C. tan⁻¹$\left( \frac{2\sigma q}{\varepsilon_{0}mg} \right)$
• D. Zero

Answer 1

Answer: (A)

tan⁻¹$\left( \frac{\sigma q}{2\varepsilon_{0}mg} \right)$

Answer 2

In equilibrium, along x-axis,

T sin θ = qE ⇒ T sin θ

= q $\frac{\sigma}{2\varepsilon_{0}}$ . . . (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 θ = $\frac{q\sigma}{2\varepsilon_{0}mg}$⇒ θ = tan⁻¹

$\left( \frac{\sigma q}{2\varepsilon_{0}mg} \right)$ ∴