Question

A charged oil drop is moving downward with a velocity $V_1$. As it acquires an additional charge it moves up with the velocity $V_2$ in the same electric field. It fall freely with a velocity $V$ in the absence of electric field. The ratio of the charges before and after acquiring additional charge is

• A. $\frac{V_1+V}{V_2-V_1}$
• B. $\frac{V_1+V_2}{2V}$
• C. $\frac{V-V_1}{V+V_2}$
• D. $\frac{V_2-2V_1}{2V_1+V}$

Answer 1

The correct option is C $\frac{V-V_1}{V+V_2}$

For the first case,

$W=6\pi \eta V_1+q_{1}E----\left(i\right)$

For the second case after gaining extra charge.

$W+6\pi \eta r{V}_2=q_{2}E----\left(ii\right)$

Also given if no electric field is applied,

$W=6\pi \eta rV----\left(iii\right)$

So, from (I),(II) and (III)

$\frac{q_1}{q_2}=\frac{V-V_1}{V+V_2}$

So, the answer is option (C).