A small bob of mass 100 mg and charge +10 is connected to an insulating string of length 1 m. It is brought near to an infinitely long non-conducting sheet of charge density ' σ ' as shown in figure. If string subtends an angle of 45° with the sheet at equilibrium the charge density of sheet will be. (Given, ε₀ = 8.85 × 10⁻¹² F/m and acceleration due to gravity, g = 10 m/s²) (a) 1.77 nC/m² (b) 0.885 nC/m² (c) 885 nC/m² (d) 17.7 nC/m²

Answer: (a). At equilibrium tan 45° = qE/mg ⇒ qE = mg = 10⁻⁴ × 10 = 10⁻³ N E = 10⁻³10⁻⁵ = 100 N/C Non-conducting sheet: E = (σ)/(2ε₀) ⇒ σ = 2 × 8.85 × 10⁻¹² × 100 = 1.77 × 10⁻⁹ C/m² So the answer is (a) 1.77 nC/m².

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Motion of Charges in an Electric Field

A small bob of mass 100 mg and charge $+10\,\mu\text{C}$ is connected to an insulating string of length 1 m. It is brought near to an infinitely long non-conducting sheet of charge density ' $\sigma$ ' as shown in figure. If string subtends an angle of $45^\circ$ with the sheet at equilibrium the charge density of sheet will be. (Given, $\epsilon_0 = 8.85 \times 10^{-12}\,\frac{\text{F}}{\text{m}}$ and acceleration due to gravity, $g = 10\,\frac{\text{m}}{\text{s}^2}$ )

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Answer: (a)

At equilibrium $\tan 45^\circ = \frac{qE}{mg} \Rightarrow qE = mg = 10^{-4} \times 10 = 10^{-3}\text{ N}$
$E = \frac{10^{-3}}{10^{-5}} = 100\text{ N/C}$
Non-conducting sheet: $E = \frac{\sigma}{2\epsilon_0} \Rightarrow \sigma = 2 \times 8.85 \times 10^{-12} \times 100 = 1.77 \times 10^{-9}\text{ C/m}^2$
So the answer is (a) $1.77\text{ nC/m}^2$ .

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