Chapter 23: Q. 60 (page 656)

An electric field can induce an electric dipole in a neutral atom or molecule by pushing the positive and negative charges in opposite directions. The dipole moment of an induced dipole is directly proportional to the electric field. That is, $\vec{p} = α \vec{E}$ , where $α$ is called the polarizability of the molecule. A bigger field stretches the molecule farther and causes a larger dipole moment.
a. What are the units of $α$ ?
b. An ion with charge $q$ is distance $r$ from a molecule with polarizability $α$ . Find an expression for the force ${\vec{F}}_{i o n o n d i p o l e}$ .

Expert verified

a. The units of $α$ is $\frac{{(C)}^{2} \times {(s)}^{2}}{k g}$

b. The required expression is : ${\vec{F}}_{i o n o n d i p o l e} = \frac{[2 α \times {(k)}^{2} \times {(q)}^{2}]}{r^{5}}$

Step by step solution

Given :

The dipole moment of an induced dipole is directly proportional to the electric field. That is : $\vec{p} = α \vec{E}$

A bigger field stretches the molecule farther and causes a larger dipole moment.

Theory used :

Unit of Dipole moment is coulomb-meter

Unit of Electric field is $m \cdot k g \cdot s^{- 3} \cdot A^{- 1}$

Solving $\vec{p} = α \vec{E}$ for units of $α$ :

localid="1649133782229" $\vec{p} = α \vec{E} \Rightarrow α = \frac{\vec{p}}{\vec{E}} = \frac{{(C)}^{2} \times {(s)}^{2}}{k g}$

On a dipole, the ion's force $F$ is:

localid="1649133791721" $F = \frac{[2 α \times {(k)}^{2} \times {(q)}^{2}]}{r^{5}}$

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