Explanation:
When the oil drop is falling freely under the effect of gravity in a viscous medium with terminal speed \(v\), then
\(mg = 6\pi \,\eta rv\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
To make the oil drop to move upward with terminal velocity \(v\) if \(E\) is the electric field intensity applied, then
\(E = mg + 6\pi \eta r = mg + mg = 2mg\)
\( \Rightarrow E = \frac{{2\,mg}}{q}\)