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The dipole moment of the water molecule (H₂O) is 6.17 × 10⁻³⁰C⋅m. Consider a water molecule located at the origin whose dipole moment [tex]\bold{p}[/tex] points in the +x-direction. A chlorine ion (Cl⁻), of charge −1.60 × 10⁻¹⁹C, is located at x = 3.00 × 10⁻⁹m. Assume that x 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.
A) Find the magnitude of the electric force that the water molecule exerts on the chlorine ion.
B) What is the direction of the electric force.
-x-direction or +x-direction
C) Is this force attractive or repulsive?

Respuesta :

Answer:

Explanation:

Electric field due to a dipole is given by the expression

E = 2P / d³ on axial line , Here P is dipole moment and d is distance of point from dipole on axial line

= 2 x 6.17 x 10⁻³⁰ / 3 x 10⁻⁹

= 4.11 x 10⁻²¹ N/C

Electric force on chlorine ion

= electric field x charge

= 4.11 x 10⁻²¹ x 1.60 × 10⁻¹⁹

= 6.576 x 10⁻⁴⁰ N

Direction of electric force is parallel to axis of dipole  in negative x - direction . This force  is Attractive .

Answer:

a) [tex]F=6.6\times 10^{-13}\ N[/tex]

B) The direction of this force is towards -x direction.

C) This force is attractive in nature.

Explanation:

Given:

  • dipole moment of water (in +x direction), [tex]p=6.17\times 10^{-30}\ C.m[/tex]
  • charge on the chlorine ion, [tex]Q_{cl}=1.6\times 10^{-19}\ C[/tex]
  • position of chlorine ion from the water molecule, [tex]x=3\times 10^{-9}\ m[/tex]

A)

Force can be given as:

[tex]F=\frac{1}{4\pi.\epsilon_0} \times \frac{2p\times Q_{cl}}{x^3}[/tex]

[tex]F=9\times 10^{9}\times \frac{2\times 6.17\times 10^{-30}\times 1.60\times 10^{-19}}{(3\times 10^{-9})^3}[/tex]

[tex]F=6.6\times 10^{-13}\ N[/tex]

B)

The direction of this force is towards -x direction i.e. attractive in nature.

C)

This force is attractive in nature.

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