Question

An oil drop carrying a charge of 4 electrons has a mass of $3.2\times {10}^{-17}$kg. It is falling freely in air with terminal speed. The electric field required to make the drop move upwards with the same speed is $(g=10m{s}^{-2})$

• A. $2\times {10}^3{Vm}^{-1}$
• B. $1\times {10}^3{Vm}^{-1}$
• C. $3\times {10}^3{Vm}^{-1}$
• D. $8\times {10}^3{Vm}^{-1}$

Answer 1

The correct option is A $2\times {10}^3{Vm}^{-1}$

We know,

Terminal velocity is achieved when Forces balance each other.

Here,

$mg=qE$

$3.2\times {10}^{-17}\times g=4\times 1.6\times {10}^{-19}\times E$

$E=2\times {10}^3$

Now,

Whether it be upwards velocity or downward velocity the condition of terminal velocity is same.

Hence, the Electric field will be same for both the cases.

Hence Option $\text{A}$ is the correct answer