A water molecule has an electric dipole moment 6.4 × 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

Question

A water molecule has an electric dipole moment 6.4 × 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 × 10-10 (B) 4 × 10-11 (C) 4 × 10-12 (D) 4 × 10-13

Tardigrade Admin · Accepted Answer

Electric dipole moment of a water molecule

p = 6.4 \times 1 0^{- 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 1 0 ^{- 30} C m}{1.6 \times 1 0 ^{- 19} C}

= 4 \times 1 0^{- 11} m

Q. A water molecule has an electric dipole moment $6.4 \times 1 0^{- 30} C m$ when it is in vapour state. The distance in metre between the centre of positive and negative charge of the molecule is

$4 \times 1 0^{- 10}$

$4 \times 1 0^{- 11}$

$4 \times 1 0^{- 12}$

$4 \times 1 0^{- 13}$

Electric dipole moment of a water molecule
$p = 6.4 \times 1 0^{- 30} C m, p = q d$
where $d$ is the distance between the centre of positive and negative charge of the molecule.
$\Rightarrow d = \frac{p}{q} = \frac{6.4 \times 1 0 ^{- 30} C m}{1.6 \times 1 0 ^{- 19} C}$
$= 4 \times 1 0^{- 11} m$

Q. A water molecule has an electric dipole moment 6.4×10−30Cm when it is in vapour state. The distance in metre between the centre of positive and negative charge of the molecule is

4×10−10

4×10−11

4×10−12

4×10−13