Question

A charged dust particle of radius $5\times {10}^{-7}$ m is located in a horizontal electric field having an intensity of $6.28\times {10}^5$v/m. The surrounding medium is air with coefficient of viscosity $\eta =1.6\times {10}^{-5}NS/m^2$. If the particle has a charge of $7.2\times {10}^{-15}$, then it moves with a uniform horizontal speed of

• A. 10
• B. 20
• C. 30
• D. 40

Answer 1

Answer: (C)

30

$E=6.28\times {10}^{5}v/m$
As force is balanced in horizontal direction,
Electrostatic force $=$ viscous force
$\Rightarrow 6.28\times {10}^{5}q=6\pi \times 1.6\times {10}^{-5}\times 5\times {10}^{-7}\times v$

$\Rightarrow \frac{6.28\times {10}^{17}}{6\pi \times 1.6\times 5}=\frac{v}{q}$

$\Rightarrow \frac{v}{q}=4.1645\times {10}^{15}$

$\therefore V=30$

So, the answer is option (C).