Q91 / 120 · VITEEE 2010 · Physics

A neutral water molecule $\left(\mathrm{H}_{2} \mathrm{O}\right)$ in its vapour state has an electric dipole moment of magnitude $6.4 \times 10^{-30} \mathrm{C}-\mathrm{m}$. How far apart are the molecules centres of positive and negative charges?

  • A
    $4 \mathrm{~m}$
  • B
    $4 \mathrm{~mm}$
  • C
    $4 \mu \mathrm{m}$
  • Correct answer
    D
    $4 \mathrm{pm}$

Solution

In a neutral water molecule, there are 10 electrons and 10 protons. So, its dipole moment $p=q(2 l)=10 e(2 l)$ Hence length of the dipole $=$ distance between centres of positive and negative charges $$ 2 l=\frac{p}{10 e}=\frac{6.4 \times 10^{-30}}{10 \times 1.6 \times 10^{-19}}=4 \times 10^{-12} \mathrm{~m} $$

Hence, the correct answer is option: 'D'.

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