Question

The figure below represents the set-up of Millikan’s oil drop experiment which was used by Millikan to determine the charge on an electron. Tiny droplets of oil in the form of mist are sprayed into the chamber P. Some of these droplets pass through the small hole in the metal plate I and are ionized by X-rays in chamber Q. 

 

If it is observed that an ionized oil droplet having a mass of 3.2 x 10^-14 kg and carrying a charge of - 6.4 x 10^-19 C, remains stationary between the metal plates I and II when a potential difference ‘V’ is applied between the plates, then 

(a) What is the direction of the applied electric field in chamber Q? Give reason. 

(b) What is the potential difference ‘V’ applied between the metal plates, if the plates are separated by a distance of 1 cm? 

(Assume there is negligible drag force experienced by the oil droplet and take g = 10m/s²) 

Answer 1

(a) The applied electric field in the chamber Q is in the downward direction, from metal plate I to metal plate II. 

The charged oil droplet will experience a downward pull due to gravity. For the droplet to be stationary an equivalent upward force should act on the oil droplet. 

This is possible only when the metal plate I acquires a positive charge and the metal plate II acquires a negative charge. 

(b) When the charged oil droplet is stationary 

qE = mg 

E = V/d 

6.4 × 10^-19 x V/10^-2 = 3.2 x 10^-14 × 10 

V = 0.5 × 10⁴ V

V = 5000 V