a) Explain what is meant by quantization of charge. (b) In a Millikan's oil drop experiment, an oil drop of weight 1.5 × 10^-14 N is held stationary between plates 10mm apart by applying a p.d. of 470V between the plates. (i) State the condition necessary for the drop to remain stationary. Also, sketch the forces acting on the oil drop. (ii) Calculate the charge on the oil-drop. (iii) Explain what would happen if the above oil drop is suddenly struck by a stray alpha particle.

Question:

a) Explain what is meant by quantization of charge.

(b) In a Millikan's oil drop experiment, an oil drop of weight 1.5 × 10^-14 N is held stationary between plates 10mm apart by applying a p.d. of 470V between the plates.

(i) State the condition necessary for the drop to remain stationary. Also, sketch the forces acting on the oil drop.

(ii) Calculate the charge on the oil-drop.

(iii) Explain what would happen if the above oil drop is suddenly struck by a stray alpha particle.

Solution:

(a) Quantization of charge means that the charge only exists in discrete values. That means the value of charge will be the integral multiple of e (1.6 x 10^-19C).

(b)

(i) Given,
Weight of drop, W = 1.5 x 10^-14 N
Distance between plates, d = 10mm = 10 x 10^-3 m
P.d., V = 470V
Since the drop is stationary between the plates, the electrical force is balanced by the weight of the drop, i.e.

\[\begin{array}{l}qE = mg\\or,q\frac{V}{d} = 1.5 \times {10^{ - 14}}\\or,{\rm{ q  =  }}\frac{{1.5 \times {{10}^{ - 14}} \times 10 \times {{10}^{ - 3}}}}{{470}} = 3.19 \times {10^{ - 19}}\end{array}\]

Thus, the charge on the drop is 3.19 x 10^-19 C.

(ii) Since oil drop is negatively and alpha particle is positively charged, they will attract each other. If F be the force of attraction between them, q₁ is the charge of oil drop and q₂ be the charge of alpha particle, then

$F = \frac{{{q_1}{q_2}}}{{4\pi { \in _0}{r^2}}}$ , Where r is distance between them.