Question

In Millikan's Experiment, the potential applied to the plates are measured as 500V, 250V, 166V and 125V so that the same drop is stationary when it is constant. Then the ratio of these charges on the drop is
A. 1:2:3:4 B. 1:2:3:5
C. 1:3:4:5 D. 1:4:5:2

Answer 1

Answer:

given =

poteintials = 750v ,250v,187.5v and 150v

to find = ratio of charge =?

solution,

force on a charge in a feild (E) = Eq

E =   V/ d

the drop is stationary ,so the upwards force on the drop due to the electric feild is equal and opposite to the weight of the drop,

vq / d = mg

vq = mgd

q ∝ 1 / v

750 : 250: 187.5:  150 ..........( multiply by 2 )

1500 :500 : 375 :300.......... (divide by 25 )

60:20: 15:12

q = 1 /v =  1/60 : 1/20 : 1/15:1/12 ...[multiply by 60 ]

60/60 : 60/20:60/14 :60/12

1:3:4:5 [charge ratio]

hence X,Y & Z

X= 1

Y= 3

Z =4

SO the correct option is 3 i.e 1:3:4:5