A neutral water molecule `(H_(2)O)` in its vapour state has an electric dipole moment of magnitudes `6.4xx10^(-30)C-m`. How far apart are the molecules centres of positive and negative charge

Options

A. `4 xx 10^(-10)`
B. `4 xx 10^(-11)`
C. `4 xx 10^(-12)`
D. `4 xx 10^(-13)`

Answer

B

Explanation

Electric dipole moment of a water molecule
`p = 6.4 xx 10^(-30)C m, p = qd`
where d is the distance between the centre of positive and negative charge of the molecule.
`rArr d = (p)/(q) = (6.4 xx 10^(-30)Cm)/(1.6 xx 10^(-19)C) = 4 xx 10^(-11)m`