Question

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
`\bold{p}` 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?

Answer 1

a)
`F=6.6* 10^(-13)\ N`

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

C) This force is attractive in nature.

Step-by-step explanation:

Given:

• dipole moment of water (in +x direction),
  `p=6.17* 10^(-30)\ C.m`

• charge on the chlorine ion,
  `Q_(cl)=1.6* 10^(-19)\ C`

• position of chlorine ion from the water molecule,
  `x=3* 10^(-9)\ m`

A)

Force can be given as:

`F=(1)/(4\pi.\epsilon_0) * (2p* Q_(cl))/(x^3)`

`F=9* 10^(9)* (2* 6.17* 10^(-30)* 1.60* 10^(-19))/((3* 10^(-9))^3)`

`F=6.6* 10^(-13)\ N`

B)

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

C)

This force is attractive in nature.

Answer 2

Step-by-step 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 .