Millikan’s oil drop experiment: task 2

The power supply to the plates is turned back on and the electric force acting on the oil droplet is:

$F_{e}=qE\; \; \; \; \; \; \; (7)$

where q is the charge on the oil droplet and E is the electric field between the plates. For parallel plates,

$E=\frac{V}{d}\; \; \; \; \; \; \; (8)$

where V is the potential difference across the plates and d is the distance between the plates. Combining eq7 and eq8,

$q=\frac{F_{e}\: d}{V}\; \; \; \; \; \; \; (9)$

If V is adjusted until the oil droplet remained stationary,

$F_{e}=mg\; \; \; \; \; \; \; (10)$

Combining eq9 and eq10,

$q=\frac{mgd}{V}$

Our objective is to determine the value of q, but it is difficult to accurately measure m. Therefore, we have to carry out another task.