The electric field of the molecule:

$\overline{)\stackrel{\mathbf{\rightharpoonup}}{\mathbf{E}}{\mathbf{=}}\frac{\stackrel{\mathbf{\rightharpoonup}}{\mathbf{p}}}{\mathbf{2}\mathbf{\pi}{\mathbf{\epsilon}}_{\mathbf{0}}{\mathbf{x}}^{\mathbf{3}}}}$

The magnitude of electric force:

$\overline{)\stackrel{\mathbf{\rightharpoonup}}{\mathbf{F}}{\mathbf{=}}{\mathbf{q}}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{E}}}$

Substituting:

The dipole moment of the water molecule (H_{2}O)is 6.17×10^{−30 }C⋅m. Consider a water molecule located at the origin whose dipole moment *p *points in the positive *x* direction. A chlorine ion (Cl^{−}), of charge −1.60×10^{−19}C, is located at *x*=3.00×10^{−9} meters. Assume that this *x* value is much larger than the separation *d* between the charges in the dipole, so that the approximate expression for the electric field along the dipole axis can be used.

Find the magnitude of the electric force, ignoring the sign, that the water molecule exerts on the chlorine ion.

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