Question

A water molecule has an electric dipole moment $6.4 \times 10^{-30} \,C m$ 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. $4 \times 10^{-13}$

Answer 1

Answer: (B)

Electric dipole moment of a water molecule
$p=6.4 \times 10^{-30} C m , p=q d$
where $d$ is the distance between the centre of positive and negative charge of the molecule.
$\Rightarrow d=\frac{p}{q}=\frac{6.4 \times 10^{-30} C m }{1.6 \times 10^{-19} C }$
$=4 \times 10^{-11} m$