Question

A charged oil drop falls with terminal velocity $V_0$ in the absence of electric field. An electric field 'E' keeps it stationary. The drop acquires additional charge q and starts moving upwards with velocity $V_0$. The initial charge on the drop was

• A. 4q
• B. 2q
• C. q
• D. q/2

Answer 1

The correct option is C q

Let the charge initially be
In 1st case,
weight $6\pi \eta r{V}_0$ ------(i)
Now when E is applied,
E$=$ Weight --------(ii)
After acquiring extra charge Q,
$\Rightarrow 6\pi \eta r{v}_0+weight=(q+Q)E$
$\Rightarrow Q=q\left[using\right(i\left)and\right(ii\left)\right]$