Question

An oil drop. carrying six electronic charge and having a mass of ${1.6\times 10}^{-12}g$, falls with some terminal velocity in a medium. What magnitude of vertical electric held is required to make the drop wove upwind with the same speed as it was formerly moving downward with? Ignore buoyancy.

• A. ${10}^5{NC}^{-1}$
• B. ${10}^4{NC}^{-1}$
• C. $3.3\times {10}^4{NC}^{-1}$
• D. $3.3\times {10}^5{NC}^{-1}$

Answer 1

The correct option is C $3.3\times {10}^4{NC}^{-1}$

$\begin{array}{cc}\text{Here}y=6e,m& =1.6\times {10}^{-12}g\\ & =1.6\times {10}^{-15}kg\end{array}$

Let the force due to airs resistance be 9 ,

In situation, when drop is falling down,

with velocity $v$

$mg=r$

Insituation, when drop is rising up with same

v, Now, the force due to air resistance will

act downwards

$mg+r=f$

Now,

$F=qE$

$mg+r=qE$

$2mg=qE[$ since, $mg$ is equal to $r]$

$E=\frac{2mg}{9}$

$E=\frac{2\times 1.6\times {10}^{-15}\times 10}{6\times 1.6\times {10}^{-19}}$

$E=3.3\times {10}^4{NC}^{-1}$