An oil drop of mass $m$ carries charge $q$. It is kept at rest in an electric field between two parallel horizontal plates separated by distance $d$, as illustrated. The potential difference across the plates is $V$ and the acceleration due to free fall is $g$. What is the charge-to-mass ratio $\frac{q}{m}$ for the oil drop?
- A$\frac{gd}{V}$
- B$\frac{V}{dg}$
- C$\frac{gV}{d}$
- D$\frac{d}{Vg}$