Question

If the shown system is in equilibrium, then the tension in the silk thread is

• A. $0.1\text{N}$
• B. $0.2\text{N}$
• C. $0.3\text{N}$
• D. $0.4\text{N}$

Answer 1

The correct option is B $0.2\text{N}$

Given,
Surface charge density of the plate, $\sigma =7.8\times {10}^{-7}{\text{C/m}}^2$
Charge on the particle, $q=4\times {10}^{-6}\text{C}$

Free body diagram of the particle:


Let $T$ be the tension acting in the silk thread.
From the equilibrium of forces along horizontal direction,
$T\sin {60}^{\circ }=qE$

$\Rightarrow T=\frac{q\sigma }{2{ϵ}_0\sin {60}^{\circ }}[\because E=\frac{\sigma }{2{ϵ}_0}]$

Substituting the values,
$\Rightarrow T=\frac{4\times {10}^{-6}\times 7.8\times {10}^{-7}}{2\times 9\times {10}^{-12}\times \sin {60}^{\circ }}$

$\therefore T=0.2\text{N}$

Hence, option (b) is the correct answer.