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 × 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 × 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, ρ = 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 E 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

  • Rupesh
  • 2019-06-15 17:23:02

Thank u sir


  • Ujwal
  • 2019-06-11 21:54:12

Nice work


  • Ujwal
  • 2019-06-11 21:53:53

Nice work


  • Atul Gupta
  • 2019-05-15 10:47:37

nice solution


  • Sudeep
  • 2019-04-08 15:36:35

The answer is 9.89m sir


  • Lolu
  • 2017-11-04 16:23:35

Lol...nice one...Rofl


  • Sivani Gupta
  • 2017-08-23 08:27:12

Good


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 2 , Question 25: An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 ×....