Question

A charged particle of mass $1.0\text{g}$ is suspended through a silk thread of length $40\text{cm}$ in a horizontal electric field of $4\times {10}^4\text{N/C}$. If the particle stays at a distance of $24\text{cm}$ from the vertical wall in equilibrium, find the charge on the particle.

• A. $1.9\times {10}^{-7}\text{C}$
• B. $2.9\times {10}^{-7}\text{C}$
• C. $3.9\times {10}^{-7}\text{C}$
• D. $4.9\times {10}^{-7}\text{C}$

Answer 1

The correct option is A $1.9\times {10}^{-7}\text{C}$


From equilibrium of forces,

$T\sin \theta =F_e$ and $T\cos \theta =mg$

$\Rightarrow \tan \theta =\frac{F_e}{mg}\dots \dots \left(1\right)$

Also, from the diagram,

$\sin \theta =\frac{24}{40}\Rightarrow \sin \theta =\frac{3}{5}\Rightarrow \theta ={37}^{\circ }\dots \dots \left(2\right)$

From (1) and (2),

$\tan {37}^{\circ }=\frac{q\times 4\times {10}^4}{1\times {10}^{-3}\times 10}$

$\Rightarrow \frac{3}{4}=\frac{q\times 4\times {10}^4}{1\times {10}^{-3}\times 10}$

$\therefore q=1.9\times {10}^{-7}\text{C}$

Hence, $\left(A\right)$ is the correct answer.