An oil drop of 12 excess electrons is held stationary under a constant electric field of $2.55 imes 10^4$ N/

Question

An oil drop of 12 excess electrons is held stationary under a constant electric field of $2.55 	imes 10^4$ N/C. The density of the oil is $1.26 	ext{ g/cm}^3$. The radius of the drop is:

Accepted Answer

For the oil drop to remain stationary, the upward electric force must balance the downward gravitational force: $qE = mg$.

1. Calculate charge $q = ne = 12 	imes 1.6 	imes 10^{-19} 	ext{ C}$.
2. Express mass in terms of density ($ho$) and radius ($r$): $m = ho 	imes \frac{4}{3}\pi r^3$.
   (Note: Convert density to SI units: $1.26 	ext{ g/cm}^3 = 1.26 	imes 10^3 	ext{ kg/m}^3$).
3. Equate forces: $qE = ho \frac{4}{3}\pi r^3 g$.
4. Solve for $r$: $r = \left( \frac{3qE}{4\pi ho g} ight)^{1/3}$.
   Substituting values: $r = \left( \frac{3 	imes 12 	imes 1.6 	imes 10^{-19} 	imes 2.55 	imes 10^4}{4 	imes 3.14 	imes 1260 	imes 9.81} ight)^{1/3} \approx 9.82 	imes 10^{-7} 	ext{ m}$.
5. Convert to mm: $9.82 	imes 10^{-7} 	ext{ m} = 9.82 	imes 10^{-4} 	ext{ mm}$. 
(See NCERT Physics Class 12, Exercise 1.25).

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