Suppose a uniformly charged wall provides a uniform electric field of 2 × 10⁴ N/C normally. A charged particle of mass 2 g being suspended through a silk thread of length 20 cm and remain stayed at a distance of 10 cm from the wall. Then the charge on the particle will be 1/√(x) where x = ___________. [use g = 10 m/s² ] Numerical answer type. Work it out, then open the answer.

Answer: 3. Thread of length 20 cm from the wall with the particle 10 cm away: sin θ = 10/20 ⇒ θ = 30°. Equilibrium: tan θ = qE/mg ⇒ q = (mgtan 30°)/(E) q = 2 × 10⁻³ × 10√(3) × 2 × 10⁴ = 1/√(3) × 10⁻⁶ C = 1/√(3) So x = 3.

Suppose a uniformly charged wall provides a uniform electric field of… | JEE Main 2024 (1 Feb, Shift 2)

Q13JEE Main 2024 · 1 FEB · SHIFT 2Medium

Motion of Charges in an Electric Field

Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^{4}\text{ N/C}$ normally. A charged particle of mass $2\text{ g}$ being suspended through a silk thread of length $20\text{ cm}$ and remain stayed at a distance of $10\text{ cm}$ from the wall.

Then the charge on the particle will be $\frac{1}{\sqrt{x}}\,\mu\text{C}$ where $x =$ ___________. [use $g = 10\text{ m/s}^2$ ]

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Thread of length 20 cm from the wall with the particle 10 cm away: $\sin\theta = \frac{10}{20} \Rightarrow \theta = 30^\circ$ .
Equilibrium: $\tan\theta = \frac{qE}{mg} \Rightarrow q = \frac{mg\tan 30^\circ}{E}$
$q = \frac{2 \times 10^{-3} \times 10}{\sqrt{3} \times 2 \times 10^{4}} = \frac{1}{\sqrt{3}} \times 10^{-6}\text{ C} = \frac{1}{\sqrt{3}}\,\mu\text{C}$
So $x = 3$ .

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