Browse by Stream
Discuss Doubts
Home
Search for Exam, Articles, Questions
get app
Go To Your Study Dashboard
  1. Home
  2. An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10^4 NC–1 (Millikan’s oil drop experiment). The density of the oil is 1.26 g cm^–3. Estimate the radius of the drop

Filters

Select Stream
Select Type
Select Exam
Select Subject
Select Chapter
Select Exercise
Q&A - Ask Doubts and Get Answers
Q
NCERT

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10^4 NC–1 (Millikan’s oil drop experiment). The density of the oil is 1.26 g cm^–3. Estimate the radius of the drop

Q 1.25: An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 \times 10^{4}NC^{-1} (Millikan’s oil drop experiment). The density of the oil is 1.26 g cm-3. Estimate the radius of the drop. 

Answers (1)
Views
H Harsh Kankaria

The force due to the electric field is balancing the weight of the oil droplet.

Weight of the oil drop = Densityx Volume of the droplet x g = \rho \times \frac{4}{3}\pi r^3 \times g

Force due to the electric field = E x q

Charge on the droplet, q = No. of excess electrons x Charge of an electron = 12\times q_{e} = 12\times 1.6\times10^{-19} = 1.92\times10^{-18} C

Balancing forces:

\rho \times \frac{4}{3}\pi r^3 \times g = E\times q

Putting known and calculated values:

 \\ r^3 =\frac{3}{4\pi} E\times q / \rho\times g = \frac{3}{4}\times\frac{7}{22}\times(2.55\times10^4)\times 1.92\times10^{-18} / (1.26\times10^3 \times10 ) \\ = 0.927\times10^{-18} m^3

r = 0.975\times10^{-6} m = 9.75\times10^{-4} mm

 

Similar Questions

Related Chapters

Exams
Articles
Questions