Question

An oil drop, carrying six electronic charges and having a mass of $1.6\times {10}^{-12}g$ , falls with some terminal velocity in a medium. What magnitude of vertical electric fields is required to make the drop move upward with the same speed as it was formerly moving downwards with? Ignore buoyancy.
A ${.10}^{5}N{C}^{-1}$
B ${.10}^{4}N{C}^{-1}$
C.$3.3\times {10}^{4}N{C}^{-1}$
D.$3.3\times {10}^{5}N{C}^{-1}$

Answer 1

Hint :In the first case only gravitational force is acting. But in the second case drag force is also acting in the downward direction so we need to put the drag force along with the gravitational force. Equating both the cases we can get the vertical electric field.
 $E={\displaystyle \frac{mg}{3e}}$ and $F=mg$ .

Complete Step By Step Answer:
In order to get the solution, first we need to take the first case;
There is only gravitational force to balance drag force
 $F=mg$
Given, mass of the oil drop is $1.6\times {10}^{-12}g$ ,
Now;
For the second case, the drop is moving upward so drag force will be in downward direction.
So we have;
 $F+mg=6eE$
Here $E$ is the vertical electric field of the oil drop.
 $e$ is the charge of the electron.
Considering both the cases we just putting the value of force in the second case, we get;
$$\begin{gathered} \Rightarrow F+mg=6eE \\ \Rightarrow mg+mg=6eE \\ \Rightarrow 2mg=6eE \\ \Rightarrow E={\displaystyle \frac{2mg}{6e}} \end{gathered}$$
So vertical electric field is $E={\displaystyle \frac{mg}{3e}}$
After putting the value of charge of electron and given mass of oil drop we get;
$$\begin{gathered} \Rightarrow E={\displaystyle \frac{mg}{3e}} \\ \Rightarrow E={\displaystyle \frac{1.6\times {10}^{-15}\times 10}{3\times 1.6\times {10}^{-19}}} \\ \Rightarrow E={\displaystyle \frac{10}{3\times {10}^{-4}}} \\ \Rightarrow E=3.3\times {10}^{4}N{C}^{-1} \end{gathered}$$
So the magnitude of the vertical electric field of the oil drop is $3.3\times {10}^{4}N{C}^{-1}$ .
Hence option D is the correct answer.

Note :
While taking the mass of the oil drop, convert it into kilograms as mass is given in grams. There is only gravitational force to balance drag force. In the second case drag force is also acting in the downward direction so we need to put the drag force along with the gravitational force.