Q. A water molecule has an electric dipole moment $6.4 \times 10^{- 30}\ cm$ when it is in vapour state. The distance in metre between the centre of positive and negative charge of the molecule is
A) $4 \times 10^{- 10}$
B) $4 \times 10^{- 11}$
C) $4 \times 10^{- 12}$
D) $4 \times 10^{- 13}$
Explanation
(b) Electric dipole moment of a water molecule
$p = 6.4 \times 10^{- 30}Cm$
$p = qd$
Where $d$ is the distance between the centre of positive and negative charge of the molecule
$\Rightarrow d = \frac{p}{q} = \frac{6.4 \times 10^{- 30}Cm}{1.6 \times 10^{- 19}C} = 4 \times 10^{- 11}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.