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:
$\begin{align}
  & (A)2\times {{10}^{3}}V{{m}^{-1}} \\
 & (B)4\times {{10}^{3}}V{{m}^{-1}} \\
 & (C)3\times {{10}^{3}}V{{m}^{-1}} \\
 & (D)8\times {{10}^{3}}V{{m}^{-1}} \\
\end{align}$

Answer 1

Hint: Under the condition of terminal velocity, the net force on the oil drop will be zero. When falling downwards, the force due to drag (friction due to air) will be equal to weight of oil drop. And when moving upwards with the same velocity, the net electric field should balance both the downward forces, that is gravity and the drag.

Complete answer:
The force $F$ on the oil drop when it is freely falling under gravity with a terminal speed of $v$ , having a radius $r$ in a fluid of viscosity $\eta $ is given by:
 $\Rightarrow F=6\pi \eta rv$
Now, since the oil drop has achieved its terminal velocity, the net force acting on it must be zero.
Therefore, the weight of oil drop will be balanced by the drag force mentioned above.
$\therefore mg=6\pi \eta rv$ [Let this expression be equation number (1)]
Now, in the second case an Electric field (say $E$ ) has been applied and finally the oil drop is moving upwards with the same speed.
Therefore, we can say that the drag force will be in the direction opposite to the velocity.
Hence, the net downward force in this case will be the sum of gravity and drag.
On balancing forces in both the direction, we get:
$\Rightarrow Eq=mg+6\pi \eta rv$
Using equation number (1), we get:
$\Rightarrow Eq=mg+mg$
$\Rightarrow Eq=2mg$
$\Rightarrow E=\dfrac{2mg}{q}$\[\]
It has been given in the problem:
$\begin{align}
  & \Rightarrow m=3.2\times {{10}^{-17}}kg \\
 & \Rightarrow q=2\times (1.6\times {{10}^{-19}})C \\
 & \Rightarrow g=10m{{s}^{-2}} \\
\end{align}$
Putting these values in the above equation and solving for E, we get:
$\begin{align}
  & \Rightarrow E=\dfrac{2\times 3.2\times {{10}^{-17}}k\times 10}{2\times (1.6\times {{10}^{-19}})}V{{m}^{-1}} \\
 & \therefore E=2\times {{10}^{3}}V{{m}^{-1}} \\
\end{align}$
Hence, the electric field required to move the oil drop upwards with the same terminal velocity is $2\times {{10}^{3}}V{{m}^{-1}}$.

Hence, option (A) is the correct option.

Note:
The drag force, that is, the force due to friction between the object and the fluid in which it is moving is always opposite to the direction of motion. This is the main reason that when the oil drop was falling, the drag force was in upwards direction but when it was moving upwards, it changed its direction to downwards direction.