In an experiment simulating Millikan's oil drop method, a charged spherical oil droplet of mass mmm and charge QQQ is observed between two horizontal metal plates. The upper plate is positive.
When the potential difference is set to zero, the droplet falls in air and reaches a constant terminal velocity v1=4.2×10−5 m s−1v_1 = 4.2 \times 10^{-5} \text{ m s}^{-1}v1=4.2×10−5 m s−1.
Given measurements:
Show that the mass mmm of the droplet is approximately 9.3×10−16 kg9.3 \times 10^{-16} \text{ kg}9.3×10−16 kg.
The potential difference across the plates is now set to V=820 VV = 820 \text{ V}V=820 V, causing the droplet to rise at a constant terminal speed v2=1.1×10−4 m s−1v_2 = 1.1 \times 10^{-4} \text{ m s}^{-1}v2=1.1×10−4 m s−1. The plate separation is d=12.0 mmd = 12.0 \text{ mm}d=12.0 mm.
Using the relation:
v2v1=VQdmg−1 \frac{v_2}{v_1} = \frac{VQ}{dmg} - 1 v1v2=dmgVQ−1Calculate the charge QQQ on the droplet, and deduce whether this value is consistent with the basic unit of charge, e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C}e=1.6×10−19 C.
