Q. A charged water drop whose radius is $0.1\ \mu m$ is in equilibrium in an electric field. If charge on it is equal to charge of an electron, then intensity of electric field will be $\left( g = 10ms^{- 1} \right)$
A) 1.61$\ N/C$
B) 26.2$\ N/C$
C) 262$\ N/C$
D) 1610$\ N/C$
Explanation
(c) In balance condition $QE = mg = \left( \frac{4}{3}\pi r^{3}\rho \right)g$
$\Rightarrow E = \frac{4 \times (3.14)\left( 0.1 \times 10^{- 6} \right)^{3} \times 10^{3} \times 10}{3 \times 1.6 \times 10^{- 19}} = 262N/C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.