Molecular $$AB$$  has a bond length of $$1.61\mathop {\text{A}}\limits^{\text{o}} $$  and a dipole moment of $$0.38\,D.$$  The fractional charge on each atom ( absolute magnitude ) is : $$\left( {{e_0} = 4.802 \times {{10}^{ - 10}}esu} \right)$$

Solution :
$$\eqalign{ & 1\,e = 1.602 \times {10^{ - 19}}C \cr & 1\,esu = 3.33 \times {10^{ - 10}}C \cr & \frac{{1\,e}}{{1\,esu}} = \frac{{1.602 \times {{10}^{ - 19}}C}}{{3.33 \times {{10}^{ - 10}}C}} \cr & 1\,e = 4.802 \times {10^{ - 10}}esu \cr & {\text{Dipole moment = }}q \times {\text{distance}} \cr & \Rightarrow 1\,D \approx {\text{1}}{{\text{0}}^{ - 18}}esu\,cm \cr & 0.38 \times {10^{ - 18}}esu\,cm = q \times \left( {1.61 \times {{10}^{ - 8}}cm} \right) \cr & q = 2.36 \times {10^{ - 11}}esu \cr & q = \frac{{2.36 \times {{10}^{ - 11}}\,esu}}{{4.802 \times {{10}^{ - 10}}esu}} \cr & q = 0.049 \cr & q \approx 0.05\,{\text{fractional charge}} \cr} $$