Q. A neutral water molecule $(H_{2}O)$ in it’s vapor state has an electric dipole moment of magnitude $6.4 \times 10^{- 30}\ C - m.$ How far apart are the molecules centres of positive and negative charge
A) $4\ m$
B) $4\ mm$
C) $4\ \mu m$
D) $4\ pm$
Explanation
(d) There are 10 electrons and 10 protons in a neutral water molecule.
So it’s dipole moment is $p = q(2l) = 10\ e\ (2l)$
Hence length of the dipole $i.e.$ distance between centres of positive and negative charges is
$2l = \frac{p}{10e} = \frac{6.4 \times 10^{- 30}}{10 \times 1.6 \times 10^{- 19}} = 4 \times 10^{- 12}m = 4\ pm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.