Question

An oil drop carrying a charge of $2$ electrons has a mass of $3.2\times 10^{- 17} \, kg$ . It is falling freely in the air with terminal speed. The electric field required to make the drop move upwards with the same speed is

• A. $2 \, \times \, 10^{3}V / m$
• B. $4 \, \times \, 10^{3}V / m$
• C. $3 \, \times \, 10^{3}V / m$
• D. $8 \, \times \, 10^{3}V / m$

Answer 1

Answer: (A)

$q=2e; \, m=3.2 \, \times \, 10^{- 17}kg$

viscous force $6\pi \eta rv=mg$ (initially) ......(i)

now motion upwards
$\therefore viscouse \, force \, downwards$
$6\pi \etarv+mg=qE\ldots \ldots \left(i i\right)$
Put value from (i) in (ii)
$qE=2mg$
$E=\frac{2 m g}{q}=\frac{2 \times 3.2 \times 1 0^{- 17} \times 10}{2 \times 1.6 \times 1 0^{- 19}}=2 \, \times \, 10^{3}V / m$