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 additonal charge q and starts moving upwards with velocity $v_0.$ The initial charge on the drop was

• A. 4q
• B. 2q
• C. q
• D. $\frac{q}{2}$

Answer 1

The correct option is D $\frac{q}{2}$

Using Stokes law, $F=6\pi \eta rv$

when drop is stationary, $q_{0}E=6\pi \eta r{v}_0$

when drop moves upward, $qE=6\pi \eta r(v_0+v_0)=2q_{0}E$

$\therefore q_0=\frac{q}{2}$

So, the answer is option (D).