Question

A group of students are recreating Millikan’s Oil Drop experiment in which a charged oil droplet is suspended in the electric field created by two horizontal charged plates. The droplet experiences a downward force due to gravity and an upward force due to the electrostatic force from the charges. The forces are balanced so that the droplet does not accelerate. In this case, the two horizontal parallel plates are 0.80 cm apart. A sphere of mass 2.73 x 10–13 kg remains stationary when the potential difference between the plates is 315 V with the upper plate negative.
Draw the FBD of the oil droplet.
Is the sphere positively or negatively charged? Explain how you know.
Calculate the magnitude of charge on the sphere.
How much excess or deficit of electrons does the sphere have?

Answer 1

Distance between horizontal plates $d =0.80 \, \text{cm}$

mass of the sphere $m =2.73\cdot 10^{-13} \, \text{kg}$

Potential difference $U = 315 \, \text{V}$

Gravity acceleration $g = 9.8 \, \text{ms}^{-2}$

Electrical field between horizontal plates:

$$E=U/d$$

Gravity force equals to electrical force, applied to the sphere:

$$mg = q E = qU /d$$

Therefore

$$q = \frac{mgd}{U} = 6.8 \cdot 10^{-17} C$$

As upper plate is negative, Electrical field directed upward

As electrical force is epplied upward, therefore q has to be positive value. It means lack (deficit) of electrons.

Elementary charge equals to

$$e = 1.6 \cdot 10^{-19}\, \text{C}$$

Number of deficit electrons:

$$N = q/e = 420$$