An oil drop of 12 excess electrons is he | Class 12 Physics Chapter Electric Charges and Field, Electric Charges and Field NCERT Solutions

Question:

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 x 104 N C-1 in Millikan's oil drop experiment. The density of the oil is 1.26 g cm-3. Estimate the radius of the drop. (g = 9.81 m s-2; e = 1.60 x 10-19 C).

Answer:

Excess electrons on an oil drop, n = 12

Electric field intensity, E = 2.55 × 104 N C - 1

Density of oil, ÃÂ = 1.26 gm/cm3 = 1.26 × 103 kg/m3

Acceleration due to gravity, g = 9.81 m s - 2

Charge on an electron, e = 1.6 × 10 - 19 C

Radius of the oil drop = r

Force (F) due to electric field is equal to the weight of the oil drop (W)

F = W

Eq = mg

Ene    

Where,

q = Net charge on the oil drop = ne

m = Mass of the oil drop

= Volume of the oil drop × Density of oil

= 9.82 × 10 - 4 mm

Therefore, the radius of the oil drop is 9.82 × 10 - 4 mm.


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Comments

  • Aqeel Ahmad
  • 2019-05-02 18:31:25

Nice work


  • Aqeel Ahmad
  • 2019-05-02 18:31:09

The question looks so difficult


Comment(s) on this Question

Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 25: An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 x 104 ....