An oil drop (density of oil = ρ), of radius r, falls vertically in the region between two plates, having a separation d and a potential difference V between them. The charge on the drop is (–q). Taking the density of air as σ and the coefficient of viscosity of air as η, the terminal velocity acquired by the drop (assuming the electrical force on it. to be smaller than the gravitational force on it), would be
(1) \([\frac{9}{2}\frac{r^2(ρ-σ)g}{η}-\frac{qV}{6\piηr}]\)
(2) \([\frac{2}{9}\frac{r^2(ρ-σ)g}{η}-\frac{qV}{6\piηr}]\)
(3) \([\frac{2}{9}\frac{r^2(ρ-σ)g}{η}-\frac{qV}{6\piηrd}]\)
(4) \([\frac{9}{2}\frac{r^2(ρ-σ)g}{η}-\frac{qV}{6\piηrd}]\)
