A water molecule has an electric dipole moment 7 × 10-30 C - m when it is in vapour state. The distance between the center of positive and negative charge of the molecule is

Question

A water molecule has an electric dipole moment 7 × 10-30 C - m when it is in vapour state. The distance between the center of positive and negative charge of the molecule is (A) 5 × 10-11 m (B) 2 × 10-11 m (C) 4.1 × 10-1 m (D) 4.375 × 10-11 m

Tardigrade Admin · Accepted Answer

P=q × 2 a Distance between the center is a P=q × a a=(7 × 10-30/1.6 × 10-19) =4.375 × 10-11 m

Q. A water molecule has an electric dipole moment $7 \times 1 0^{- 30} C - m$ when it is in vapour state. The distance between the center of positive and negative charge of the molecule is

$5 \times 1 0^{- 11} m$

$2 \times 1 0^{- 11} m$

$4.1 \times 1 0^{- 1} m$

$4.375 \times 1 0^{- 11} m$

$P = q \times 2 a$
Distance between the center is a
$P = q \times a$
$a = \frac{7 \times 1 0 ^{- 30}}{1.6 \times 1 0 ^{- 19}}$
$= 4.375 \times 1 0^{- 11} m$

Q. A water molecule has an electric dipole moment 7×10−30C−m when it is in vapour state. The distance between the center of positive and negative charge of the molecule is

5×10−11m

2×10−11m

4.1×10−1m

4.375×10−11m